Sequence approaching infimum

Question

Given the fact that for a non-empty subset $K ⊂ V$ and $J : V → \mathbb R$ the number $\inf_{v \in K} J(v)$ exists, why does a minimizing sequence always exist? So I read somewhere this: A minimizing sequence of a function $J$ on the set $K$ is a sequence $(u_n)_{n∈ \mathbb N}$ such that $u_n \in K$ for all $n$ and $\lim_{n\to\infty}J(u_n) = \inf_{v∈K} J(v)$. By definition of the infimum value of $J$ on $K$ there always exist minimizing sequences! It seems like a logical consequence but I don't get it...

Answer 1

Part of the reason you might be having trouble "getting" this result is that it's impossible to visualize! (At least, it is impossible to visualize unless you have more information about $V$ and $K$ than you've given us.) The result tells you that the sequence exists, but it doesn't give you any way of knowing what the points of the sequence actually are, nor any particular means of choosing them.

It turns out that the quoted result (as written) is equivalent to the

Axiom of Countable Partial Choice: Given a sequence of non-empty sets $(A_n)_{n\in\Bbb N},$ there is an infinite subset $I$ of $\Bbb N$ and a function $f:I\to\bigcup_{n\in\Bbb N}A_n$ such that $f(n)\in A_n$ for all $n\in I.$

Note that the above result tells us little about the sets $A_n,$ the set $I,$ and the function $f.$ So, while we might be able to grasp this result conceptually, we wouldn't necessarily be able to take a given sequence of sets $A_n$ and determine what $I$ and $f$ satisfy the desired properties. Let's prove the equivalence!

On the one hand, suppose the minimizing sequence result holds, and that we have a sequence $(A_n)_{n\in\Bbb N}$ of non-empty sets. Let $$V=K=\bigcup_{n\in\Bbb N}A_n\times\{n\}.$$ Note that the elements of $K$ have the form $\langle a,n\rangle,$ where $a\in A_n\subseteq A$ and $n\in\Bbb N.$ Define $J:V\to\Bbb R$ by $J(a,n)=\frac1{n+1}.$ Since the sets $A_n$ are non-empty, then so are the sets $A_n\times\{n\}.$ Hence, for each $n\in\Bbb N,$ there is some $\langle a,n\rangle\in A_n\times\{n\},$ meaning that $J(a,n)=\frac1{n+1}.$ Thus, the range of $J$ is $\left\{\frac1{n+1}:n\in\Bbb N\right\}.$ Readily, the infimum of this set is $0.$ By the desired result, there is a minimizing sequence--that is, a sequence $\bigl(\langle a_k,n_k\rangle\bigr)_{k\in\Bbb N}$ of points of $K$ such that $\lim_{k\to\infty}J(a_k,n_k)=0.$ In other words, $\lim_{k\to\infty}\frac1{n_k+1}=0.$ Now, if the sequence $(n_k)_{k\in\Bbb N}$ took on only finitely-many values, then we couldn't have $\lim_{k\to\infty}\frac1{n_k+1}=0.$ Hence, $(n_k)_{k\in\Bbb N}$ takes on infinitely-many values. Letting $I=\{n\in\Bbb N:n=n_k\text{ for some }k\in\Bbb N\},$ we therefore have that $I$ is an infinite subset of $\Bbb N.$ Given any $n\in I,$ there is some least $k\in\Bbb N$ such that $n=n_k;$ and given that least $k,$ we define $f(n)=a_k.$ Readily, then, we can show that $f:I\to \bigcup_{n\in\Bbb N}A_n,$ and $f(n)\in A_n$ for all $n\in I,$ so ACPC holds.

On the other hand, suppose ACPC holds, and suppose that $K$ is a nonempty subset of $V$ and $J:V\to\Bbb R$ a function such that $\inf_{v\in K}J(v)$ exists. Let $x=\inf_{v\in K}J(v),$ and for each $n\in\Bbb N,$ let $$A_n:=\left\{v\in K:J(v)<x+\frac1{n+1}\right\}.$$ By definition of $x,$ we can readily show that each $A_n$ is non-empty. Hence, there is an infinite subset $I$ of $\Bbb N$ and a function $f:I\to\bigcup_{n\in\Bbb N}A_n$ such that $f(n)\in A_n$ for all $n\in I.$ Since $I$ is an infinite subset of $\Bbb N,$ then there is a function $h:\Bbb N\to I$ sending the least element of $\Bbb N$ to the least element of $I$, the second-least element of $\Bbb N$ to the second-least element of $I$, and so on. Finally, letting $v_n=f\bigl(h(n)\bigr)$ for all $n\in\Bbb N,$ we find that $(v_n)_{n\in\Bbb N}$ is a sequence of points of $K,$ and we can prove by construction that $\lim_{n\to\infty}J(v_n)=x=\inf_{v\in K}J(v),$ as desired.

Answer 2

Hint:

Gap Lemma: Let $\varnothing \neq A \subseteq \mathbb R$ and $x = \inf A$. Given any $\varepsilon > 0$, there is an $a \in A$ such that $a - x < \varepsilon$.

Then, we can use the gap lemma with $\varepsilon_n = \frac{1}{n}$ to get a sequence $(a_n)_{n \in \mathbb N}$ such that $\lim_{n\to\infty} a_n = x$.