Building neighborhoods around arbitrary real numbers that will contain "minimal" number of elements from the sequence $a_n=n\sin \frac{\pi n}{4}$

Question

Suppose we are given the sequence $\{a_n\}_{n=1}^\infty$ defined by the rule $\forall n\in\mathbb{N}, a_n=n\sin \frac{\pi n}{4}$.

We can show that this is equivalent to:

$\forall n\in\mathbb{N},a_n= \begin{cases} \frac{n}{\sqrt{2}} & n \equiv 1\pmod{8} \lor n \equiv 3\pmod{8} \\ n & n \equiv 2\pmod{8} \\ 0 & n \equiv 0\pmod{8}\lor n \equiv 4\pmod{8} \\ -\frac{n}{\sqrt{2}} & n \equiv 5\pmod{8} \lor n \equiv 7\pmod{8} \\ -n & n\equiv 6 \pmod{8} \end{cases} $

Now the problem is to prove\show that around each number we take $L\in\mathbb{R}$ we can create\build a neighborhood such that:

1. If $L$ does not equal to any element in the sequence $\{a_n\}_{n=1}^\infty$, Then we can build a neighborhood of $L$ such that no other element from the sequence will reside in it.

2. If $L$ equals some element in the sequence $\{a_n\}_{n=1}^\infty$, Then, If $L$ equals zero (as zero is an element in the sequence), Then we can build a neighborhood of $L$ that will contain an infinite number of elements from the sequence, And If $L$ does not equals zero, Then we can build a neighborhood of $L$ that will contain exactly one element from the sequence.

Translating the problem into logic/set-theory notation we get that we have to show:

1. If $\forall n\in\mathbb{N}, L\neq a_n$, Then $\exists \epsilon\in (0,\infty), |\{n\in\mathbb{N}|a_n\in N_\epsilon(L)\}|=0$

2. If $\exists m\in\mathbb{N}, L= a_m$, Then,

   If $L=0$, Then $\exists \epsilon\in (0,\infty), |\{n\in\mathbb{N}|a_n\in N_\epsilon(L)\}|=\aleph_0$,

   And If $L\neq 0$, Then $\exists \epsilon\in (0,\infty), |\{n\in\mathbb{N}|a_n\in N_\epsilon(L)\}|=1$

Note: The notation $N_\epsilon (L)$ denotes the neighborhood of $L$ with radius $\epsilon$, I.e. $N_\epsilon(L)=(L-\epsilon,L+\epsilon)$

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My try for proving 1:

Since in this case it is given that $\forall n\in\mathbb{N}, L\neq a_n$ we can conclude that $\forall n\in\mathbb{N}, a_n-L\neq 0$ and thus $\forall n\in\mathbb{N}, | a_n-L|>0$, Now If we define the set $A=\{|a_n-L| | n\in\mathbb{N}\}$ we get that this set is a non-empty set of real numbers that is bounded below by $0$ and thus its infimum $\inf(A)=\inf\limits_{n\in\mathbb{N}} |a_n-L|$ exists and satisfies $\inf\limits_{n\in\mathbb{N}}|a_n-L|\geq 0$ (as $0$ is a lower bound of $A$ and thus must be less than or equal to its infimum), Now if we just show that $\inf\limits_{n\in\mathbb{N}}|a_n-L|>0$ we could choose $\epsilon=\inf\limits_{n\in\mathbb{N}}|a_n-L|\in (0,\infty)$ and we would get that $\forall n\in\mathbb{N}, a_n\notin N_\epsilon(L)$ because if we suppose by contradiction that $\exists n\in\mathbb{N}, a_n\in N_\epsilon(L)$, Then we would get by the fact that $|a_n-L|\in A$ and by definition of infimum that $|a_n-L|\geq \inf(A)=\inf\limits_{n\in\mathbb{N}}|a_n-L|=\epsilon$ which contradicts the fact that $a_n\in N_\epsilon(L)$ (which is equivalent to $|a_n-L|<\epsilon$), Thus it must be the case that $\forall n\in\mathbb{N},a_n\notin N_\epsilon(L)$ and we can conclude that $\{n\in\mathbb{N}|a_n\in N_\epsilon(L)\}=\emptyset$ and thus $|\{n\in\mathbb{N}|a_n\in N_\epsilon(L)\}|=0$ as was to be shown.

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My try for proving 2:

Since in this case it is given that $\exists m\in\mathbb{N}, L = a_m$, And since $0$ is in the image of $\{a_n\}_{n=1}^\infty$ (i.e. $0$ is an element in the set $\{a_n|n\in\mathbb{N}\}$) we get that there are two possibilities: $L=0$ or $L\neq 0$

If $L=0$, Any $\epsilon$ we choose from the set $(0,\infty)$ will work, As we can define the set $T=\{n\in\mathbb{N}|n\equiv 0 \pmod{8}\}$ and it is clear that for this set we have $\forall n\in T, a_n\in N_\epsilon(L)$ [because $\forall n\in T, a_n=0$ and because $0\in N_\epsilon(0)=N_\epsilon(L)$], And therefore we get that $T\subseteq \{n\in\mathbb{N}|a_n\in N_\epsilon(L)\}\subseteq \mathbb{N}$, But since $|T| = |\mathbb{N}| =\aleph_0$ we get that it must be the case that $|\{n\in\mathbb{N}|a_n\in N_\epsilon(L)\}|=\aleph_0$ as was to be shown.

Now if $L\neq 0$, We can define the set $A=\{|a_n-L| | m\neq n\in\mathbb{N}\}$, And we get that this set is a non-empty set of real numbers that is bounded below by $0$ and thus its infimum $\inf(A)=\inf\limits_{m\neq n\in\mathbb{N}} |a_n-L|$ exists and satisfies $\inf\limits_{m\neq n\in\mathbb{N}}|a_n-L|\geq 0$ (as $0$ is a lower bound of $A$ and thus must be less than or equal to its infimum), Now if we just show that $\inf\limits_{m\neq n\in\mathbb{N}}|a_n-L|>0$ we could choose $\epsilon=\inf\limits_{m\neq n\in\mathbb{N}}|a_n-L|\in (0,\infty)$ and we would get that $\forall m\neq n\in\mathbb{N}, a_n\notin N_\epsilon(L)$ because if we suppose by contradiction that $\exists m\neq n\in\mathbb{N}, a_n\in N_\epsilon(L)$, Then we would get by the fact that $|a_n-L|\in A$ and by definition of infimum that $|a_n-L|\geq \inf(A)=\inf\limits_{m\neq n\in\mathbb{N}}|a_n-L|=\epsilon$ which contradicts the fact that $a_n\in N_\epsilon(L)$ (which is equivalent to $|a_n-L|<\epsilon$), Thus it must be the case that $\forall m\neq n\in\mathbb{N},a_n\notin N_\epsilon(L)$ and we can conclude that $\{n\in\mathbb{N}|a_n\in N_\epsilon(L)\}=\{m\}$ (because $L=a_m$) and thus $|\{n\in\mathbb{N}|a_n\in N_\epsilon(L)\}|=1$ as was to be shown.

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Thanks for any hint\help on how to prove that those infimums are indeed positive....

Answer 1

Proof of Case (1):

We will show that indeed $\inf\limits_{n\in\mathbb{N}}|a_n-L|>0$:

Because $\forall n\in\mathbb{N}, L\neq a_n$, We get that in particular $L\neq 0$ and thus we have two possibilities: $L>0$ or $L<0$

If $L>0$, Then we can define an increasing sequence of natural numbers $\{n_k\}_{k=1}^\infty$ as follows: $\forall k\in\mathbb{N}, n_k := 8k+1$, And thus we can consider the subsequence $\{a_{n_k}\}_{k=1}^\infty$ of the sequence $\{a_n\}_{n=1}^\infty$ that corrsepsonds to this index sequence $\{n_k\}_{k=1}^\infty$, Now since $\forall k\in\mathbb{N}, n_k\equiv 1\pmod{8}$, We can conclude by definition of the sequence $\{a_n\}_{n=1}^\infty$ that $\forall k\in\mathbb{N}, a_{n_k}=\frac{n_k}{\sqrt{2}}$.

Now since it is clear that $\lim\limits_{k\to \infty} (8k+1)=\infty$, And since $\frac{1}{\sqrt{2}}>0$ we can conclude that $\lim\limits_{k\to\infty}\frac{8k+1}{\sqrt{2}}=\infty$, And thus $\lim\limits_{k\to\infty}\frac{n_k}{\sqrt{2}}=\infty$ which implies that $\lim\limits_{k\to\infty}a_{n_k}=\infty$.

Now by definition of limit at $\infty$, We get that $\exists K\in\mathbb{N},\forall K<k\in\mathbb{N},L<a_{n_k}$, Now if we choose any $k_0\in\mathbb{N}$ that satisfies $K<k_0$ we get that for this particular $k_0$ we have $L<a_{n_{k_0}}$.

Now let’s define $N:={n_{k_0}}+8$ (We add this $8$ inorder to ensure that $L$ will be an element in the set $B$ that will be defined below very soon), It is clear that $N\in\mathbb{N}$.

We will show that for this particular $N$ we have $N\equiv 1\pmod{8}$, $9\leq N$ and $L<a_N=\frac{N}{\sqrt{2}}$:

Since $n_{k_0}\equiv 1\pmod{8}$ and since $8\equiv 0\pmod{8}$ we get that $n_{k_0}+8\equiv 1\pmod{8}$, And therefore $N\equiv 1\pmod{8}$ as was to be shown.

Now since $n_{k_0}\in\mathbb{N}$, We get that $1\leq n_{k_0}$, And thus $9\leq n_{k_0}+8$, Which implies that $9\leq N$ as was to be shown.

Now by definition of the sequence $\{a_n\}_{n=1}^\infty$ and by using the fact $N\equiv 1\pmod{8}$ that we’ve just shown, We get that $a_N=\frac{N}{\sqrt{2}}$, Now since $0<8$, We get $n_{k_0}<n_{k_0}+8$, But because $N=n_{k_0}+8$, We conclude $n_{k_0}<N$, Now by dividing this inequality by $\sqrt{2}>0$ we get $\frac{n_{k_0}}{\sqrt{2}}<\frac{N}{\sqrt{2}}$, But since $a_{n_{k_0}}=\frac{n_{k_0}}{\sqrt{2}}$ and since $a_N=\frac{N}{\sqrt{2}}$, We can conclude that $a_{n_{k_0}}<a_N$, Now since $L<a_{n_{k_0}}$ we can conclude by transitivity of the $<$ relation that $L<a_N$ as was to be shown.

Now let’s define a set $B:=\{|a_n-L||n\in\{1,...,N\}\}$, Since $B$ is a non-empty finite set of real numbers we get that its minimum $\min(B)=\min\limits_{n\in\{1,...,N\}}|a_n-L|$ exists and is an element in $B$, I.e. $\min(B)=\min\limits_{n\in\{1,...,N\}}|a_n-L|\in B$, Now since $\forall n\in\mathbb{N}, |a_n-L|>0$ we get that in particular $\forall n\in\{1,...,N\},|a_n-L|>0$, Which implies that $\forall b\in B, b>0$, And thus, In particular for $\min(B)=\min\limits_{n\in\{1,...,N\}}|a_n-L|\in B$ we have $\min(B)=\min\limits_{n\in\{1,...,N\}}|a_n-L|>0$.

Now we’ll show that $L\in B$:

Since $4\leq 9$ and since $9\leq N$, We get that $4\leq N$, And thus $4\in\{1,...,N\}$, And we can conclude that $|a_4-L|\in B$, But since $4\equiv 4\pmod{8}$, We get $a_4=0$, And thus $|a_4-L|=|0-L|=|-L|=|L|$, And we conclude that $|L|\in B$, But because $L>0$ we get that $|L|=L$, And so $L\in B$ as was to be shown.

Now we’ll prove that $\forall n\in\mathbb{N}, \min\limits_{n\in\{1,...,N\}}|a_n-L|\leq |a_n-L|$:

Let $n\in\mathbb{N}$, Then there are two cases: $n\leq N$ or $N<n$

If $n\leq N$, Then we get that $n\in\{1,...,N\}$, And thus $|a_n-L|\in B$, And we can conclude that $\min\limits_{n\in\{1,...,N\}}|a_n-L|=\min(B)\leq |a_n-L|$.

If $N<n$, Then there are five cases: $a_n=0$, $a_n=\frac{n}{\sqrt{2}}$, $a_n=n$, $a_n=-\frac{n}{\sqrt{2}}$ or $a_n=-n$

If $a_n=0$, Then we get that $|a_n-L|=|0-L|=|-L|=|L|=L$, And as we’ve already shown that $L\in B$, We can conclude that $|a_n-L|\in B$, And thus $\min\limits_{n\in\{1,...,N\}}|a_n-L|=\min(B)\leq |a_n-L|$.

If $a_n=\frac{n}{\sqrt{2}}$, Then because $N<n$, We get by dividing this inequality by $\sqrt{2}>0$, That $\frac{N}{\sqrt{2}}<\frac{n}{\sqrt{2}}$, But since $a_N=\frac{N}{\sqrt{2}}$ we conclude that $a_N<a_n$, And thus (#) $a_N-L<a_n-L$, But because $L<a_N$ we get that $0<a_N-L$, Which implies that $0<a_n-L$, And thus $|a_N-L|=a_N-L$ And $|a_n-L|=a_n-L$ and we can conclude by (#) that $|a_N-L|<|a_n-L|$, Now because $|a_N-L|\in B$, We get that $\min\limits_{n\in\{1,...,N\}}|a_n-L|=\min(B)\leq |a_N-L|$, And we conclude that $\min\limits_{n\in\{1,...,N\}}|a_n-L|<|a_n-L|$.

If $a_n=n$, Then because $1<\sqrt{2}$, We get $\frac{1}{\sqrt{2}}<1$, And by multiplying this inequality by $N>0$, We get $\frac{N}{\sqrt{2}}<N$, But since $N<n$, We get $\frac{N}{\sqrt{2}}<n$, But since $a_N=\frac{N}{\sqrt{2}}$ we conclude that $a_N<a_n$, And thus (#) $a_N-L<a_n-L$, But because $L<a_N$ we get that $0<a_N-L$, Which implies that $0<a_n-L$, And thus $|a_N-L|=a_N-L$ And $|a_n-L|=a_n-L$ and we can conclude by (#) that $|a_N-L|<|a_n-L|$, Now because $|a_N-L|\in B$, We get that $\min\limits_{n\in\{1,...,N\}}|a_n-L|=\min(B)\leq |a_N-L|$, And we conclude that $\min\limits_{n\in\{1,...,N\}}|a_n-L|<|a_n-L|$.

If $a_n=-\frac{n}{\sqrt{2}}$, Then we get $|a_n-L|=|-\frac{n}{\sqrt{2}}-L|=|\frac{n}{\sqrt{2}}+L|$, But since $n>0$, We get that $\frac{n}{\sqrt{2}}>0$, And since $L>0$, We can conclude that $\frac{n}{\sqrt{2}}+L>0$, And so $|\frac{n}{\sqrt{2}}+L|=\frac{n}{\sqrt{2}}+L$, And we can conclude that $|a_n-L|=\frac{n}{\sqrt{2}}+L$, But since $\frac{n}{\sqrt{2}}>0$, We get that $\frac{n}{\sqrt{2}}+L>L$, And we can conclude that $|a_n-L|>L$, But since $L\in B$, We get that $\min\limits_{n\in\{1,...,N\}}|a_n-L|=\min(B)\leq L$, And thus $\min\limits_{n\in\{1,...,N\}}|a_n-L|< |a_n-L|$.

If $a_n=-n$, Then we get $|a_n-L|=|-n-L|=|n+L|$, But since $n>0$ and $L>0$, We get that $n+L>0$, And so $|n+L|=n+L$, And we can conclude that $|a_n-L|=n+L$, But since $n>0$, We get that $n+L>L$, And we can conclude that $|a_n-L|>L$, But since $L\in B$, We get that $\min\limits_{n\in\{1,...,N\}}|a_n-L|=\min(B)\leq L$, And thus $\min\limits_{n\in\{1,...,N\}}|a_n-L|< |a_n-L|$.

Thus we’ve shown that in each of those five cases we have $\min\limits_{n\in\{1,...,N\}}|a_n-L|\leq |a_n-L|$ as was to be shown.

Now we’ll prove that $\inf\limits_{n\in\mathbb{N}} |a_n-L|=\min\limits_{n\in\{1,...,N\}}|a_n-L|$ :

Since $\min\limits_{n\in\{1,...,N\}}|a_n-L|\in B$ and since $B\subseteq A$, We get that $\min\limits_{n\in\{1,...,N\}}|a_n-L|\in A$, But because what we’ve just shown, I.e. that $\forall n\in\mathbb{N}, \min\limits_{n\in\{1,...,N\}}|a_n-L|\leq |a_n-L|$, We conclude that $ \min\limits_{n\in\{1,...,N\}}|a_n-L|$ is the minimum element of $A$, I.e. $A$ has a minimum and its minimum satisfies $\min\limits_{n\in\mathbb{N}}|a_n-L|=\min(A)=\min\limits_{n\in\{1,...,N\}}|a_n-L|$, But because we’ve already shown that $\min\limits_{n\in\{1,...,N\}}|a_n-L|>0$, We conclude that $\min\limits_{n\in\mathbb{N}}|a_n-L|>0$, But because $\min\limits_{n\in\mathbb{N}}|a_n-L|=\inf\limits_{n\in\mathbb{N}}|a_n-L|$, We conclude that $\inf\limits_{n\in\mathbb{N}}|a_n-L|>0$ as was to be shown.

The case for $L<0$ can be proved in a similar way.

Q.E.D.

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Proof of Case (2):

(Note: we consider only the case of $L\neq 0$ as we’ve already dealt with the case $L=0$ in the text of the question itself)

We will show that indeed $\inf\limits_{m\neq n\in\mathbb{N}}|a_n-L|>0$:

First we’ll prove that $\forall m\neq n\in\mathbb{N}, L\neq a_n$:

Suppose by contradiction that $\exists m\neq n\in\mathbb{N}, L = a_n$, Since $L=a_m$, We get that $a_n=a_m$, And thus by definition of the sequence $\{a_n\}_{n=1}^\infty$, We get that (#) $n \sin\frac{\pi n}{4}=m\sin\frac{\pi m}{4}$, Now since in this case $L\neq 0$ and since $L=a_n$ , We get that $a_n\neq 0$, And thus $n\sin\frac{\pi n}{4}\neq 0$, And we conclude that it must be the case that $\sin\frac{\pi n}{4}\neq 0$, Also since $m\in\mathbb{N}$ we get that $m\neq 0$, And we can conclude that $m \sin \frac{\pi n}{4}\neq 0$, Now by dividing the equality (#) by this non-zero number we get $\frac{n \sin\frac{\pi n}{4}}{m \sin \frac{\pi n}{4}}=\frac{m\sin\frac{\pi m}{4}}{m \sin \frac{\pi n}{4}}$ Which is equivalent to (##) $\frac{n}{m}=\frac{\sin \frac{\pi m}{4}}{\sin \frac{\pi n}{4}}$, Now since $\forall q\in\mathbb{N}, \sin\frac{\pi q}{4}=0,\pm 1,\pm \frac{1}{\sqrt{2}}$, We have in particular that $\sin\frac{\pi m}{4}=0,\pm 1,\pm \frac{1}{\sqrt{2}}$ and $\sin\frac{\pi n}{4}=0,\pm 1,\pm \frac{1}{\sqrt{2}}$, But since we’ve already shown that $\sin\frac{\pi n}{4}\neq 0$, We get that $\sin\frac{\pi n}{4}=\pm 1,\pm \frac{1}{\sqrt{2}}$, Now we’ll show that $\sin\frac{\pi m}{4}\neq 0$, Because If we suppose by contradiction that indeed $\sin\frac{\pi m}{4}=0$ , Then we get by equation (##) that $\frac{n}{m}=\frac{0}{sin\frac{\pi n}{4}} = 0$ and thus $n=m\times 0=0$ which contradicts the fact that $n\in\mathbb{N}$, Thus it must be the case that $\sin\frac{\pi m}{4}\neq 0$, And we conclude that $\sin\frac{\pi m}{4}=\pm 1,\pm \frac{1}{\sqrt{2}}$, And therefore $\frac{\sin\frac{\pi m}{4}}{\sin\frac{\pi n}{4}}=\frac{\pm 1}{\pm 1},\frac{\pm 1}{\pm\frac{1}{\sqrt{2}}},\frac{\pm\frac{1}{\sqrt{2}}}{\pm 1},\frac{\pm\frac{1}{\sqrt{2}}}{\pm\frac{1}{\sqrt{2}}}=\pm 1, \pm \sqrt{2}, \pm \frac{1}{\sqrt{2}}$, And by equality (##) we conclude that $\frac{n}{m}=\pm 1,\pm \sqrt{2},\pm \frac{1}{\sqrt{2}}$, Now since $m,n\in\mathbb{N}$ we get that $\frac{n}{m}$ must be a rational number, I.e. $\frac{n}{m}\in\mathbb{Q}$, And thus it must be the case that $\frac{n}{m}\neq \pm \sqrt{2},\pm\frac{1}{\sqrt{2}}$, Also we’ll show that $\frac{n}{m}\neq -1$, Because if we suppose by contradiction that indeed $\frac{n}{m}=-1$, Then we would get that $n=-m$, But since $n\in\mathbb{N}$ we get that $0<n$ and thus $0<-m$ which implies that $m<0$, But since $m\in\mathbb{N}$ we get that $0<m$ and therefore we got thatA $0<m<0$ which implies that $0<0$ and we’ve reached a contradiction, Thus it must be the case that $\frac{n}{m}\neq -1$.

In summary we’ve shown that it must be the case that $\frac{n}{m}=\pm 1,\pm \sqrt{2},\pm \frac{1}{\sqrt{2}}$, And that it must be the case that $\frac{n}{m}\neq -1,\pm \sqrt{2},\pm \frac{1}{\sqrt{2}}$, And therefore we conclude that it must be the case that $\frac{n}{m}=1$ and we conclude that $m=n$, But this contradicts the fact that $m\neq n$, And thus it must be the case that $\forall m\neq n\in\mathbb{N}, L\neq a_n$ as was to be shown.

Now since $L\neq 0$ we get that there are two possibilities: $L>0$ or $L<0$

If $L>0$, Then we can define an increasing sequence of natural numbers $\{n_k\}_{k=1}^\infty$ as follows: $\forall k\in\mathbb{N}, n_k := 8k+1$, And thus we can consider the subsequence $\{a_{n_k}\}_{k=1}^\infty$ of the sequence $\{a_n\}_{n=1}^\infty$ that corrsepsonds to this index sequence $\{n_k\}_{k=1}^\infty$, Now since $\forall k\in\mathbb{N}, n_k\equiv 1\pmod{8}$, We can conclude by definition of the sequence $\{a_n\}_{n=1}^\infty$ that $\forall k\in\mathbb{N}, a_{n_k}=\frac{n_k}{\sqrt{2}}$.

Now since it is clear that $\lim\limits_{k\to \infty} (8k+1)=\infty$, And since $\frac{1}{\sqrt{2}}>0$ we can conclude that $\lim\limits_{k\to\infty}\frac{8k+1}{\sqrt{2}}=\infty$, And thus $\lim\limits_{k\to\infty}\frac{n_k}{\sqrt{2}}=\infty$ which implies that $\lim\limits_{k\to\infty}a_{n_k}=\infty$.

Now by definition of limit at $\infty$, We get that $\exists K\in\mathbb{N},\forall K<k\in\mathbb{N},L<a_{n_k}$, Now if we choose any $k_0\in\mathbb{N}$ that satisfies $K<k_0$ we get that for this particular $k_0$ we have $L<a_{n_{k_0}}$.

Now let’s define $M:=m+9-m\mod 8$, We’ll show that $m<M\in\mathbb{N}$ and that $M\mod 8=1$:

Since $0\leq m\mod 8\leq 7$, We get that $-7\leq -m\mod 8\leq 0$, And by adding $9$ we get $2=9-7\leq 9-m\mod 8\leq 9+0=9$, Now by adding $m$ we get $m+2\leq m+9-m\mod 8\leq m+9$, And thus $m+2\leq M\leq m+9$, But since $0<2$ we get that $m<m+2$, And we can conclude that $m<M$, Also since $m,9,m\mod 8\in\mathbb{Z}$ we conclude that $m+9-m\mod 8\in\mathbb{Z}$ and thus $M\in\mathbb{Z}$, Now since $m\in\mathbb{N}$ we get that $0<m$, And by the fact $m<M$ that we’ve just shown, We conclude that $0<M$, And thus $M\in\mathbb{N}$, And in summary we’ve got that $m<M\in\mathbb{N}$ as was to be shown.

Now we have:

(###) $M\mod 8 = (m+9-m\mod 8)\mod 8=((m+9)\mod 8- (m \mod 8)\mod 8)\mod 8=((m\mod 8+9\mod 8)\mod 8 - (m\mod 8))\mod 8=((m\mod 8+1)\mod 8-m\mod 8)\mod 8$,

And now there are two possibilities $0\leq m\mod 8\leq 6$ or $m\mod 8=7$:

If $0\leq m\mod 8\leq 6$, Then $1\leq m\mod 8+1\leq 7$, And thus $(m\mod 8+1)\mod 8=m\mod 8+1$, And we conclude by (###) that $M\mod 8=(m\mod 8+ 1-m\mod 8)\mod 8=1\mod 8=1$.

If $m\mod 8=7$, Then $M\mod 8= ((7+1)\mod 8-7)\mod 8=(8\mod 8-7)\mod 8=(0-7)\mod 8 =(-7)\mod 8=1$.

Thus we see that it is always the case that $M\mod 8=1$ as was to be shown.

Now let’s define $N:=\max (M,{n_{k_0}})+8$ (We add this $8$ inorder to ensure that $L$ will be an element in the set $B$ that will be defined below very soon), Since $M,n_{k_0}\in\mathbb{N}$, We get that $\max(M,n_{k_0})\in\mathbb{N}$, And since $8\in\mathbb{N}$, We can conclude that $\max(M,n_{k_0})+8\in\mathbb{N}$, And thus $N\in\mathbb{N}$.

We will show that for this particular $N$ we have $N\equiv 1\pmod{8}$, $9\leq N$, $L<a_N=\frac{N}{\sqrt{2}}$ and $m<N$:

Since $M\equiv 1\pmod{8}$, And Since $n_{k_0}\equiv 1\pmod{8}$, We get that $\max(M,n_{k_0})\equiv 1\pmod{8}$, Also since $8\equiv 0\pmod{8}$ we get that $\max(M,n_{k_0})+8\equiv 1\pmod{8}$, And therefore $N\equiv 1\pmod{8}$ as was to be shown.

Now since $\max(M,n_{k_0})\in\mathbb{N}$, We get that $1\leq \max (M,n_{k_0})$, And thus $9\leq \max (M,n_{k_0})+8$, Which implies that $9\leq N$ as was to be shown.

Now by definition of the sequence $\{a_n\}_{n=1}^\infty$ and by using the fact $N\equiv 1\pmod{8}$ that we’ve just shown, We get that $a_N=\frac{N}{\sqrt{2}}$, Now since $0<8$, We get $\max (M,n_{k_0})<\max (M,n_{k_0})+8$, But because $N=\max (M,n_{k_0})+8$, We conclude $\max(M,n_{k_0})<N$, But Since $n_{k_0}\leq \max (M,n_{k_0})$, We get that $n_{k_0}<N$, Now by dividing this inequality by $\sqrt{2}>0$ we get $\frac{n_{k_0}}{\sqrt{2}}<\frac{N}{\sqrt{2}}$, But since $a_{n_{k_0}}=\frac{n_{k_0}}{\sqrt{2}}$ and since $a_N=\frac{N}{\sqrt{2}}$, We can conclude that $a_{n_{k_0}}<a_N$, Now since $L<a_{n_{k_0}}$ we can conclude that $L<a_N$ as was to be shown.

Now since $0<8$, We get $\max (M,n_{k_0})<\max (M,n_{k_0})+8$, But because $N=\max (M,n_{k_0})+8$, We conclude $\max(M,n_{k_0})<N$, But since $M\leq \max (M,n_{k_0})$, We get that $M<N$, But because we’ve already shown that $m<M$ we can conclude that $m<N$ as was to be shown.

Now let’s define a set $B:=\{|a_n-L||n\in\{1,...,N\}\backslash\{m\}\}$, We’ll show that $L\in B$ and that $|a_N-L|\in B$:

Since $4\leq 9$ and since $9\leq N$, We get that $4\leq N$, And thus $4\in\{1,...,N\}$, Now we’ll show that $4\notin\{m\}$, Because if we suppose by contradiction that $4\in\{m\}$, Then we’ll get that $4=m$, And thus $a_m=a_4$, But since $4\equiv 4 \pmod{8}$, We get by definition of the sequence $\{a_n\}_{n=1}^\infty$ that $a_4=0$, And thus $a_m=0$, But since $a_m=L$ we get that $L=0$ which contradicts the fact that $L\neq 0$, And thus it must be the case that $4\notin\{m\}$, And we can conclude that $4\in\{1,...,N\}\backslash\{m\}$, And therefore we can conclude that $|a_4-L|\in B$, But since $|a_4-L|=|0-L|=|-L|=|L|$, We conclude that $|L|\in B$, But because $L>0$ we get that $|L|=L$, And so $L\in B$ as was to be shown

Also since $N\in\{1,...,N\}$ And since $N\notin\{m\}$, Because if we suppose by contradiction that $N\in\{m\}$, Then we’ll get that $N=m$, But as we’ve already shown that $m<N$, We conclude that $m<m$ and we’ve reached a contradiction, Thus it must be the case that $N\notin\{m\}$, And we can conclude that $N\in\{1,...,N\}\backslash \{m\}$, And thus $|a_N-L|\in B$ as was to be shown.

Now since $B$ is a non-empty (as we’ve just shown that $L\in B$) finite set of real numbers we get that its minimum $\min(B)=\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|$ exists and is an element in $B$, I.e. $\min(B)=\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|\in B$, Now since we’ve shown that $\forall m\neq n\in\mathbb{N}, a_n\neq L$, We conclude that $\forall m\neq n\in\mathbb{N}, a_n-L\neq 0$, Which implies that $\forall m\neq n\in\mathbb{N}, |a_n-L|>0$, And thus we get that in particular $\forall n\in\{1,...,N\}\backslash\{m\},|a_n-L|>0$, Which implies that $\forall b\in B, b>0$, And thus, In particular for $\min(B)=\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|\in B$ we have $\min(B)=\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|>0$.

Now we’ll prove that $\forall m\neq n\in\mathbb{N}, \min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|\leq |a_n-L|$:

Let $m\neq n\in\mathbb{N}$, Then there are two cases: $n\leq N$ or $N<n$

If $n\leq N$, Then we get that $n\in\{1,...,N\}$, But as $n\neq m$, We get that $n\notin \{m\}$, And thus $n\in\{1,...,N\}\backslash\{m\}$, And we conclude that $|a_n-L|\in B$, And therefore we can conclude that $\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|=\min(B)\leq |a_n-L|$.

If $N<n$, Then there are five cases: $a_n=0$, $a_n=\frac{n}{\sqrt{2}}$, $a_n=n$, $a_n=-\frac{n}{\sqrt{2}}$ or $a_n=-n$

If $a_n=0$, Then we get that $|a_n-L|=|0-L|=|-L|=|L|=L$, And as we’ve already shown that $L\in B$, We can conclude that $|a_n-L|\in B$, And thus $\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|=\min(B)\leq |a_n-L|$.

If $a_n=\frac{n}{\sqrt{2}}$, Then because $N<n$, We get by dividing this inequality by $\sqrt{2}>0$, That $\frac{N}{\sqrt{2}}<\frac{n}{\sqrt{2}}$, But since $a_N=\frac{N}{\sqrt{2}}$, We conclude that $a_N<a_n$, And thus (§) $a_N-L<a_n-L$, But because $L<a_N$, We get that $0<a_N-L$, Which implies that $0<a_n-L$, And thus $|a_N-L|=a_N-L$ and $|a_n-L|=a_n-L$, And we can conclude by (§) that $|a_N-L|<|a_n-L|$, Now because we’ve already shown that $|a_N-L|\in B$, We get that $\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|=\min(B)\leq |a_N-L|$, And we conclude that $\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|<|a_n-L|$.

If $a_n=n$, Then because $1<\sqrt{2}$, We get $\frac{1}{\sqrt{2}}<1$, And by multiplying this inequality by $N>0$, We get $\frac{N}{\sqrt{2}}<N$, But since $N<n$, We get $\frac{N}{\sqrt{2}}<n$, But since $a_N=\frac{N}{\sqrt{2}}$, We conclude that $a_N<a_n$, And thus (§) $a_N-L<a_n-L$, But because $L<a_N$, We get that $0<a_N-L$, Which implies that $0<a_n-L$, And thus $|a_N-L|=a_N-L$ and $|a_n-L|=a_n-L$, And we can conclude by (§) that $|a_N-L|<|a_n-L|$, Now because we’ve already shown that $|a_N-L|\in B$, We get that $\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|=\min(B)\leq |a_N-L|$, And we conclude that $\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|<|a_n-L|$.

If $a_n=-\frac{n}{\sqrt{2}}$, Then we get $|a_n-L|=|-\frac{n}{\sqrt{2}}-L|=|\frac{n}{\sqrt{2}}+L|$, But since $n>0$, We get that $\frac{n}{\sqrt{2}}>0$, And since $L>0$, We can conclude that $\frac{n}{\sqrt{2}}+L>0$, And so $|\frac{n}{\sqrt{2}}+L|=\frac{n}{\sqrt{2}}+L$, And we can conclude that $|a_n-L|=\frac{n}{\sqrt{2}}+L$, But since $\frac{n}{\sqrt{2}}>0$, We get that $\frac{n}{\sqrt{2}}+L>L$, And we can conclude that $|a_n-L|>L$, But since $L\in B$, We get that $\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|=\min(B)\leq L$, And thus $\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|< |a_n-L|$.

If $a_n=-n$, Then we get $|a_n-L|=|-n-L|=|n+L|$, But since $n>0$ and $L>0$, We get that $n+L>0$, And so $|n+L|=n+L$, And we can conclude that $|a_n-L|=n+L$, But since $n>0$, We get that $n+L>L$, And we can conclude that $|a_n-L|>L$, But since $L\in B$, We get that $\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|=\min(B)\leq L$, And thus $\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|< |a_n-L|$.

Thus we’ve shown that in each of those five cases we have $\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|\leq |a_n-L|$ as was to be shown.

Now we’ll prove that $\inf\limits_{m\neq n\in\mathbb{N}} |a_n-L|=\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|$ :

Since $\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|\in B$ and since $B\subseteq A$, We get that $\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|\in A$, But because what we’ve just shown, I.e. that $\forall m\neq n\in\mathbb{N}, \min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|\leq |a_n-L|$, We conclude that $ \min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|$ is the minimum element of $A$, I.e. $A$ has a minimum and its minimum satisfies $\min\limits_{m\neq n\in\mathbb{N}}|a_n-L|=\min(A)=\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|$, But because we’ve already shown that $\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|>0$, We conclude that $\min\limits_{m\neq n\in\mathbb{N}}|a_n-L|>0$, But because $\min\limits_{m\neq n\in\mathbb{N}}|a_n-L|=\inf\limits_{m\neq n\in\mathbb{N}}|a_n-L|$, We conclude that $\inf\limits_{m\neq n\in\mathbb{N}}|a_n-L|>0$ as was to be shown.

The case for $L<0$ can be proved in a similar way.

Q.E.D.

---

Practically Calculating the Infimums:

If we want to practically calculate the infimums $\inf\limits_{n\in\mathbb{N}}|a_n-L|$ and $\inf\limits_{m\neq n\in\mathbb{N}}|a_n-L|$ that we got in cases (1) and (2) (only the case in which $L\neq 0$) respectively, Then we will make use of the facts that in case (1) we have $\inf\limits_{n\in\mathbb{N}} |a_n-L|=\min\limits_{n\in\{1,...,N\}}|a_n-L|$ and in case (2) (only the case in which $L\neq 0$) we have $\inf\limits_{m\neq n\in\mathbb{N}} |a_n-L|=\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|$, Therefore inorder to get the value of the appropriate infimum we will just calculate the value of $\min\limits_{n\in\{1,...,N\}}|a_n-L|$ or $\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|$ depending on wether we are dealing with case (1) or case (2) (only the case in which $L\neq 0$).

We will consider only the case of $L>0$:

First let’s define $max\_index := \lceil\sqrt{2}L\rceil$, We’ll show that $max\_index\in\mathbb{N}$:

Since $L>0$ and $\sqrt{2}>0$, We get that $\sqrt{2}L>0$, And thus $0<\lceil \sqrt{2}L\rceil\in\mathbb{Z}$, And therefore we can conclude that $\lceil \sqrt{2}L\rceil\in\mathbb{N}$, And thus $max\_index\in\mathbb{N}$ as was to be shown.

Now we’ll prove that (#) $\forall max\_index < n\in\mathbb{N}, L\neq a_n$:

Let $max\_index <n\in\mathbb{N}$, Then by definition of $max\_index$, We get that $\lceil\sqrt{2}L\rceil<n$, But since $\sqrt{2}L\leq\lceil\sqrt{2}L\rceil$, We conclude that $\sqrt{2}L<n$, And thus (§) $L<\frac{n}{\sqrt{2}}$.

Now there are five cases for $a_n$: $a_n=0$, $a_n=\frac{n}{\sqrt{2}}$, $a_n=n$, $a_n=-\frac{n}{\sqrt{2}}$ or $a_n=-n$

If $a_n=0$, Then we can conclude by the fact $L>0$ that $L>a_n$, And thus in particular $L\neq a_n$.

If $a_n=\frac{n}{\sqrt{2}}$, Then we can conclude by (§) that $L<\frac{n}{\sqrt{2}}$, And thus $L<a_n$, And we get that in particular $L\neq a_n$.

If $a_n=n$, Then we can conclude by (§) that $L<\frac{n}{\sqrt{2}}$, But since $1<\sqrt{2}$, We get $\frac{1}{\sqrt{2}}<1$, And by multiplying this inequality by $n>0$ (as $n\in\mathbb{N}$), We get that $\frac{n}{\sqrt{2}}<n$, And thus we can conclude that $L<n$, Which implies that $L<a_n$, And thus in particular $L\neq a_n$.

If $a_n=-\frac{n}{\sqrt{2}}$, Then we get by the fact $0<\frac{n}{\sqrt{2}}$ that $-\frac{n}{\sqrt{2}}<0$, And thus $a_n<0$, But since $0<L$, We conclude that $a_n<L$, And thus in particular we have $L\neq a_n$.

If $a_n=-n$, Then we get by the fact $0<n$ that $-n<0$, And thus $a_n<0$, But since $0<L$, We conclude that $a_n<L$, And thus in particular we have $L\neq a_n$.

Thus we’ve shown that in each of those five cases we have $L\neq a_n$ as was to be shown.

By (#) we conclude that in order for us to check wether $\exists m\in\mathbb{N}, L=a_m$, It is enough for us to check wether $\exists m\in\{1,...,max\_index\},L=a_m$.

By taking this observation into account, We can program an algorithm that will determine if $L$ equals an element in the sequence $\{a_n\}_{n=1}^\infty$, And if it is, It will return the index of the first occurrence of $L$ in the sequence $\{a_n\}_{n=1}^\infty$, And if not it will return an index of $-1$, I.e. a non-existing index that will indicate that $L$ does not equal any element of the sequence $\{a_n\}_{n=1}^\infty$.

// Checks wether the argument equals an element in the sequence,
// If it is, The function will return the index in which it was found,
// And if it was not, It will return -1.

int search_for(double L)
{
    // The maximal index that must be examined:
    int max_index = ceil(sqrt(2) * L);

    // The index at which it was found (if was found):
    int index = 1;

    // Indicates wether the value was found at the current interation:
    bool is_found = false;

    do
    {
        if(a(index) == L)
            is_found = true;
        else
            index++;
    }
    while (!is_found  && index <= max_index);

    if (!is_found)
        index = -1;

    return index;
}

Now, We will use this algorithm in order to check wether $L$ equals an element in the sequence, If it is not, Then we are dealing with case (1) and the following algorithm will return the value of $\min\limits_{n\in\{1,...,N\}}|a_n-L|$, And if it is, Then we are dealing with case (2) (with $L\neq 0$) And the following algorithm will return the value of $\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|$, And thus, We will get the exact values of the infimums.

double infimum(double L)
{
   double minimal_distance; // Will hold in the end the value of the appropriate infimum.
   int m; // Holds the first index for which $L=a_m$ if any.
   int k0; // Holds the index that its existence was guaranteed in cases (1) and (2).
   int nk0; // Holds the element of the sequence $\{n_k\}$ at the index $k_0$. 
   int M; // Holds the value of $M$ as defined in case (2) [Used only if we are dealing with case (2)].
   int N; // Holds the greatest index that must be considered inorder to evaluate the minimums.

   // We have to choose an index $k_0$ that will ensure that $L<a_{n_{k_0}}$, 
   // And the $k_0$ defined this way will satisfy it.
   k0 = ceil((sqrt(2)*L-1)/8)+1;

   // The $k_0$’s element of the sequence $\{n_k\}$.
   nk0 = 8*k0+1;

   // Searches for the first index for which $L=a_m$ if any.
   m = search_for(L);

   // L equals some element in the sequence and thus $m>0$:
   if (m > 0)
   {
       // Since we are dealing here with case (2), 
       // We will define $M$ and $N$ as they were defined in case (2).
       M = m + 9 - m%8;
       N = max(M,nk0)+8;
   }

   // L does not equal any element in the sequence and thus $m=-1$.
   else
   {
       // Since we are dealing here with case (1),
       // We will define $N$ as it was defined in case (1).
       N = nk0 + 8;
   }

   // The following chunk of code will calculate the appropriate minimum based   
   // on wether we are dealing with case (1) or case (2).
   // Note:
   // If we are dealing with case (1), Then we know that $m=-1$, 
   // And thus the condition $i\neq m$ in the for loop header will never be satisfied,
   // And in the end $minimal_distance$ will hold the value $\min\limits_{n\in\{1,...,N\}}|a_n-L|$
   // And if we are dealing with case (2), Then we know that $m\in\{1,...,N\}$,
   // And thus the condition $i\neq m$ in the for loop header will skip over the
   // index $m$, And in the end $minimal_distance$ will hold the value  $\min\limits_{n\in\{1,...,N\}\backslash\{m\}}|a_n-L|$ 
   minimal_distance = INFINITY;
   for (int i = 1; i != m && i <= N; i++)
   {
       double current_distance = abs(a(i) - L);
       if (current_distance < minimal_distance)
           minimal_distance = current_distance;
   }

   return minimal_distance;
}

Note: The algorithms work only for the case $L>0$ and for the case $L<0$ some modifications are needed inorder for the algorithm to work in this case.

By putting the whole program together we get:

https://github.com/MathNerd/Projects/blob/master/calculate_infimums.c

Answer 2

Hints:

If $L$ is not an element of the sequence then there exists unique $n \in \mathbb{Z}$ such that $n < L < n+1$ and unique $m \in \mathbb{Z}$ such that $\frac{m}{\sqrt2} < L < \frac{m+1}{\sqrt2}$.

Therefore let $\varepsilon = \min\left\{\left|L-(n+1 )\right|, \left|L-n\right|, \left|L-\frac{m+1}{\sqrt2}\right|, \left|L - \frac{m}{\sqrt2}\right|\right\}$ and consider the neighbourhood $(L-\varepsilon, L + \varepsilon)$.

If $L = 0$ then any neighbourhood of $0$ will do, because $0$ appears infinitely many times in the sequence.

If $L$ appears in the sequence but $L \ne 0$, then:

• if $L \in \mathbb{Z}$ then there exists a unique $m \in \mathbb{Z}$ such that $\frac{m}{\sqrt2} < L < \frac{m+1}{\sqrt2}$. Let $\varepsilon = \min\left\{1, \left|L-\frac{m+1}{\sqrt2}\right|, \left|L - \frac{m}{\sqrt2}\right|\right\}$.

• if $L = \frac{n}{\sqrt2}$ for some $n\in \mathbb{Z}$ then there exists a unique $m \in \mathbb{Z}$ such that $m < L < m+1$. Let $\varepsilon = \min\left\{\frac1{\sqrt2}, \left|L-(m+1)\right|, \left|L - m\right|\right\}$.

and consider the neighbourhood $(L-\varepsilon, L + \varepsilon)$.