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Let the non-compact $\;E\subset \mathcal X\;$ where $\;(\mathcal X,d_0)\;$ is a metric space and denote the distance of a point $\;x\in \mathcal X\;$ from $\;E\;$ as:

$\;d(x,E)=\inf_{y\in E\;}\{d_0(x,y)\}\;$.

After some research on the Internet I found that in general, if I would like to show that $\;\inf_{y\in E\;}\{d_0(x,y)\}=d_0(x,y^*)\;$ for some $\;y^* \in E\;$, then all I have to do is to take a sequence $\;y_n \in E\;$ such that $\;d(x,y_n) \to \inf_{\;y\in E\;}\{d_0(x,y)\}\;$ and find a subsequence $\;y_{n_k}\;$ of $\;y_n\;$ with $\;y_{n_k} \to y^*\in E\;$.

Why is the above sufficient? Is it related to the sequential compactness of $\;E\;$? But if it does, how is it possible since $\;E\;$ is non-compact? Did I miss some steps in the process of minimizing the infimum distance?

I would really appreciate if somebody could help me understand how this kind of exercises work and provide me (if possible) some Analysis Theorems that I might miss.

Thanks in advance!

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The general approach you refer to applies to finding a solution $y^* \in E$ for any set $E$ and $x \in \mathcal X$ of the problem:

$\qquad d_0(x,y^*) = \inf_{y \in E}d_0(x,y)$

This approach relies purely on the fact that $d_0: \mathcal X^2 \to \Bbb R_{\ge 0}$ is a continuous function. Because $y_{n_k} \to y^*$, it follows that $d(x,y_{n_k})\to d(x,y^*)$.

What it does not do is establishing tot $y^*$ actually exists (even in $\mathcal X$), or is a part of $E$.

For the first, we need the metric space $\mathcal X$ to be complete (Cauchy sequences converge in $\mathcal X$). For the second, we can generally only establish $y^* \in E$ if $E$ is a closed set.

Since both these conditions were not given, we cannot ensure that this process of finding $y^*$ will actually manage to find an element of $E$ with the desired property.

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  • $\begingroup$ I 'm about to ask something really silly but I will. So the fact that $\;d(x,y_{n_k})\to d(x,y^*)\;$ and $\;d(x,y_n)\;$ is also convergent implies that the limit of $\;d(x,y_n)\;$ must be the same with its subsequence and hence the infimum becomes minimum, right? Let me ask you one more thing, if $\;E\;$ included functions which they had an exponential estimate, then would it be possible for $\;E\;$ to have some kind of compactness? However thanks for your answer! $\endgroup$ Jan 13 '18 at 12:45
  • $\begingroup$ You're welcome. As to the first question: indeed. As to the second, I'm afraid I don't understand it :/. As it seems unrelated, maybe it's good to ask a new question regarding that. If you do, please also include all the information you have. There are now several gaps in your description that make it hard to answer your question in full detail. $\endgroup$
    – Lord_Farin
    Jan 13 '18 at 13:07
  • $\begingroup$ You're right. It is quite general what I've asked. I'll come back with a new question for better results $\endgroup$ Jan 13 '18 at 13:24
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Set $A_x\overset{_{\mathrm{def}}}{=}\{ \mathrm{d}(x,y) : y\in E\}$. By infimum definition, we have that for all $\epsilon>0$ there is a $y_\epsilon\in E$ such that $$ \inf A_x\leq d(x,y_\epsilon)<\inf A_x+\epsilon $$
Then for all $\epsilon_n=\frac{1}{n}$ there is a $y_n\in E$ such that $$ \inf A_x\leq d(x,y_n)<\inf A_x+\frac{1}{n} $$ The point here is that if the sequence $\{y_n\}_{n\in\mathbb{N}}$ is convergent (or has a convergent subsequence ) for a point $y^\ast\in E$ we have $\inf A_x= d(x,y^\ast)$.

Now your questions:

Why is the above sufficient? Is it related to the sequentially compactness of E?

The sequential compactness of $E$ is sufficient.

But if it does, how is it possible since EE is non-compact?

Always some primitive notion of sequential compactness will be needed as a sufficient condition. For example, relatively compact subspace is sucfficient.

Did I miss some steps in the process of minimizing the infimum distance?

Having some notion of compactness, there is no loss in the constructive process of the sequence.

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  • $\begingroup$ Excuse me, but while I was reading your answer again, I had trouble understanding why the condition of a relatively compact subspace is also sufficient. If such a condition holds, then there exists a subsequence in $\;E\;$ which has a convergent subsequence in $\;\mathcal X\;$. How does this imply that inf=min? Thanks in any case $\endgroup$ Jan 17 '18 at 17:43
  • $\begingroup$ @kaithkolesidou, Yes this is true. $\endgroup$ Jan 18 '18 at 0:07

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