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Let $(X, d)$ be any arbitrary metric space. Now fix $y \in X$ and define $d_y: X \rightarrow \mathbb{R}, d_y(x)= d(x,y)$.

How can I prove that $d_y$ is continuous on $X$? (Given any fixed $y \in X$).

Also, let $A \subseteq X$ and fix $x \in X$, define $\rho(x, A) = \inf \{d(x,a) : a \in A \}$,

Show that the closure of $A$ is $\overline{A}=\{x \in X: \rho(x, A)=0 \}$

Finally, let $A \subseteq X$ be a compact subset of $X$.

Prove that $\rho(x, A) = d(x, a)$ for some $a \in A$.

TeTs
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1 Answers1

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Problem 1

Let $\varepsilon > 0$ be arbitrary, choose $\delta = \epsilon$, try to show that $$ d(x,z)<\delta\implies \lvert d_y(x) - d_y(z)\rvert < \varepsilon $$ on your own.

Problem 2

Recall that the closure of $A$ is the union of $A$ and its limit points. This means that we need to show that if a point $x$ is a limit point of $A$, i.e. if there exists a sequence $(a_n)_n\subseteq A$ that converges to $x$, then $\rho(x,A) = 0$, and that if $\rho(x,A) = 0$, then we can find a sequence $(a_n)_n\subseteq A$ that converges to $x$.

To show the first direction, show that you can always find some $a\in A$ that is arbitrarily close to $x$ by using the given sequence $(a_n)_n$ that converges to $x$. To show the second, use the definition of infimum to try to show the existence of $a\in A$ that are arbitrarily close to $x$.

Problem 3

In metric spaces, compactness is the same as sequential compactness, meaning that every sequence in $A$ has a convergent subsequence that converges inside $A$. First show that you can construct a sequence $(a_n)_n\subseteq A$ so that $d(x,a_n)$ converges to $\rho(x,A)$. Then, show that this sequence $(a_n)_n$ is convergent (hint: compact spaces are complete). Finally, show that $a\in A$, the limit of $a_n$, satisfies $\rho(x,A) = d(x,a)$.

Good luck!

Tai
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  • Thanks, is my attempt at Problem 3 correct as follows? Fix $x \in X$ and define the function $d_x: A \rightarrow \mathbb{R}, d_x(a) = d(x, a)$. By the first problem, this function is continuous on $A$ (for fixed $x \in X)$. So by the Extreme Value Theorem (we can apply it here since $A$ is compact), there exists $a^* \in A$ such that $d_x(a^*) = \inf_{a \in A} d_x(a) = \rho(x, A)$. – TeTs Jul 26 '17 at 14:44
  • Sounds good! But are you sure you are allowed to use EVT? Seems like you haven't proven it yet, given the nature of the problems. – Tai Jul 26 '17 at 16:51