Trying to understand sequence convergence proof visually

Question

Lemma:

Let $(a_n)$ and $(b_n)$ be sequences in $\mathbb{R}$ with $a_n\to a$ and $b_n\to b$. Then, for every $\varepsilon >0$, there is a $N\in > \mathbb{N}$ such that $|a_n-a|<\varepsilon$ and $|b_n-b|<\varepsilon$ for all $n\geq N$.

This basically means that we can (for a given $\varepsilon$) find an $N$ such that it "works" for $(a_n)$ and $(b_n)$ simultaneously.

Proof for the Lemma:

Let $\varepsilon >0$. Since $a_n\to a$ we can choose a $N_1\in \mathbb{N}$ with $|a_n-a|<\varepsilon$ with $n\geq N_1$. Similarly, we can choose a $N_2\in \mathbb{N}$ with $|b_n-b|<\varepsilon$ for all $n\geq N_2$. Choose $N_0:=\max \{n_1, n_2\}$, then we have $|a_n-a|<\varepsilon$ and $|b_n-b|<\varepsilon$ for all $n\geq N_0$.

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Im trying to understand this proof visually now. That's an image i drew with desmos. The black dots are $a_n:=1/n$ and the blue dots $b_=-0.25$ How would we choose $N_0$ this time? I can't really grasp the idea that this works... I guess that it means that after a certain element of the sequence they both are in the same epsilon-neighbourhood.

Answer 1

Because $(b_n)$ is constant, any $N_0$ that works for $(a_n)$ works, trivially, for $(b_n)$ too.

Notice though that generally and in your case the two sequences won't be within the same $\varepsilon$-neighbourhood for small $\varepsilon$: instead, both sequences will be within one $\varepsilon$ of their respective limits.

Take $N_0 = 2\varepsilon$, for example.

Answer 2

The number $N_0$ depends upon $\varepsilon$. For a given $\varepsilon>0$, take $N_0$ such that $N_0>\frac1\varepsilon$. Then, if $n\geqslant N_0$,$$0<\frac1n\leqslant\frac1{N_0}<\frac1\varepsilon.$$So, $n\geqslant N_0\implies\lvert a_n-0\rvert<\varepsilon$. And, of course, for every $n\in\mathbb N$, $\left\lvert b_n-\left(-\frac14\right)\right\rvert=0<\varepsilon$.

Answer 3

Choose $\varepsilon>0$ arbitrarily. The limit of $(a_n)$ is 0 so $\lvert a_n -0 \rvert=\frac{1}{n}<\varepsilon$ for $n\geq\lceil\frac{1}{\varepsilon}\rceil$. Similarly $\lvert b_n-0.25\rvert=0<\varepsilon$ for $n\geq1$. Take the maximum of both: $n\geq\lceil\frac{1}{\varepsilon}\rceil$.

Answer 4

I believe in your proof you actually want $N_0:=\max\{N_1,N_2\}$.

You know that $a_n$ can lie within distance $\epsilon$ of $a$ so long as $n$ is greater than some $N_1$. Similarly we know that $b_n$ can lie within $\epsilon$ of $b$ so long as we choose $n$ greater than some $N_2$. Both statements will be simultaneously true when we choose values of $n$ greater than whichever of $N_1$ and $N_2$ is largest.

Formally, given $\epsilon>0$, we know that there exists $N_1$ such that whenever $n>N_1$, $|a_n-a|<\epsilon$. Given again that same $\epsilon>0$, there also exists some $N_2$ such that whenever $n>N_2$ we have $|b_n-b|<\epsilon$. This is all from the definition of a limit.

We then choose $N_0:=\max\{N_1,N_2\}$. A priori we do not know the relative values of $N_1$ and $N_2$, but for argument's sake say $N_1>N_2$ so that $N_0=N_1$. Then in order for both $|a_n-a|<\epsilon$ and $|b_n-b|<\epsilon$ simultaneously, we consider values of $n>N_0=N_1$. For values of $n>N_0$, clearly $|a_n-a|<\epsilon$ by definition. Since $|b_n-b|<\epsilon$ for values of $n>N_2$, it is equally true for values of $n>N_0=N_1$ since $N_0=N_1>N_2$ and so for values of $n>N_0$ we have both statements true simultaneously.

Answer 5

When $x\in U\subseteq \Bbb R$ and $(x-r,x+r)\subseteq U$ for some $r>0$ we say that $U$ is a neighborhood (nbhd) of $x.$

One way to $define$ $x=\lim_{n\to \infty} x_n$ is that whenever $U$ is any nbhd of $x$ then $\{n\in \Bbb N: x_n\not \in U\}$ is finite.

Intuitively, imagine a sequence of discs centered at $x,$ of ever-smaller radii. No matter how small the radius is, the disc contains $x_n$ for all but finitely many $n\in \Bbb N$ (which I picture as a handful of $n$ at most, because any finite amount is tiny compared to the set of all the other $n$ such that $x_n$ $is$ in the disc.)

So if $a_n\to a$ and $b_n\to b$ and you are given some (any) $\epsilon >0,$ consider the nbhd $U(a)=(a-\epsilon,a+\epsilon)$ of $a$ and the nbhd $U(b)= (b-\epsilon,b+\epsilon)$ of $b.$

The sets $P=\{n\in \Bbb N:a_n\not \in U(a)\}$ and $Q=\{n\in \Bbb N:b_n\not \in U(b)\} $are finite.

So $P\cup Q$ is a finite subset of $\Bbb N.$ So there exists $m\in \Bbb N$ such that whenever $m\le n\in \Bbb N,$ we must have $n\not \in P\cup Q,$ that is, $n\not \in P$ and $n\not \in Q.$ This implies that $a_n\in U(a)$ and $b_n\in U(b)$ whenever $m\le n\in \Bbb N$.