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I'm studying a proof I learned in class and I don't quite understand this statement.

Let $X$ be a metric space and $C \subset Z$ a nonempty closed set. For each $x \in X$ define $f_{c}(x)=$ inf $\{ d (x,y)\mid y \in C\}$ (this is the distance from $x$ to $C$).

We want to prove that $f_{c}(x)=0$ if and only if $x\in C$.

So first let $f_{c}(x)=0$. The proof goes one to say that this fact implies that for all $n \geq 1$ there exists $z_n \in C$ such that $d(x,z_n)< \frac{1}{n}$ by the definition of inf.

I'm just a little confused on how the definition on inf gives us this implication.

I understand that $f_c(x)=0$ means that the distance from $x$ to $C$ is 0. So the inf (or least upper bound) of the distance from a point in $X$ and a point in $C$ is 0.

But I don't understand why that implies there is some sequence in $C$ where the distance from the point in $X$ and the sequence is less than $\frac{1}{n}$. Why does the definition of inf imply this?

Michael Hardy
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user185698
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2 Answers2

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Hint: $l = \inf C \iff \forall \epsilon > 0 \, \exists c \in C : c < l + \epsilon$, now consider $\epsilon = \frac{1}{n}$ for $n \in \mathbb{N}$

Edit: I just realized you might not have been given this as a definition for $\inf$, if this is the case let me know what your definition is and I'll connect it to the one I used.

DanZimm
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  • Okay, so that's just the definition of what it means to be an inf. Then set $\epsilon = \frac{1}{n}$, but that was kind of arbitrary correct? We didn't have to set it to that? So, if we let $c= z_{n}$ and we are given that $l$ = 0, then our definition becomes for all $\frac{1}{n} > 0$ there exists $z_{n} \in C$ such that $z_{n} < 0 + \frac{1}{n}$ for $n \in N$. But I guess I'm still a little confused as to why we picked a sequence in $C$. – user185698 Oct 26 '14 at 23:46
  • So the distance from our sequence in $X$ to our point in $X$ is arbitrarily small. Thus, we have that $d(x, z_{n})< \frac{1}{n}$. And so then the proof says that $z_{n} \rightarrow x$? Because that's what it means to converge right? – user185698 Oct 26 '14 at 23:51
  • Yes, that specific $\epsilon$ is rather arbitrary. The important thing about the $\epsilon$s is that they get closer and closer to $0$. We might as well have used $\frac{1}{2^n}$, or $\frac{1}{n^3+ 5n -2}$. – Arthur Oct 26 '14 at 23:54
  • Okay, thank you. And why did we choose a sequence $z_{n} \in C$ rather than a point $z \in C$? – user185698 Oct 26 '14 at 23:55
  • Because we are not guaranteed that such a point exists. If $f_C(x)=0$, we only know that no matter how close up to $x$ you go, there is a $z_n\in C$ that is closer, but from the $\inf$-definition alone we do not know that there is a point $z\in C$ with distance $0$ to $x$. In fact, there are $x$ for which this fails when $C$ is not closed. – Arthur Oct 26 '14 at 23:59
  • Also, $f_C(x)=0$ is equivalent to saying "there is a sequence in $C$ converging to $x$". Therefore, saying "Any point $x$ with $f_C(x)=0$ is contained in $C$" is the same as saying "Any converging sequence of points from $C$ converges to a point in $C$", which is the definition of $C$ being closed. – Arthur Oct 27 '14 at 00:10
  • If another part of the problem asked to show by example that $z_{n}$ need not be unique, what kind of example could I give? – user185698 Oct 27 '14 at 02:19
  • @Arthur thanks for clarifying these things – DanZimm Oct 27 '14 at 03:48
  • @user185698 do you understand what has been explained thus far? If so then just pick a closed set in $\mathbb{R}$ and pick a point outside the set and find two sequences that approach the distance inside the relevant set of distances. – DanZimm Oct 27 '14 at 03:51
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Assume that $\inf\{d(x,y):y \in C\} = 0$. Let $\epsilon = \frac{1}{2^n}$. Now $\forall n \in \mathbb{N}, \exists y_n \in C $ such that $d(x,y_n) < \frac{1}{2^n}$. Now $\displaystyle \frac{1}{2^n} \to 0 $ as $ n \to \infty$. Thus there exists a sequence in $C$ which converges to $x$. Moreover, $C$ is assumed to be closed thus the limit point $x$ is contained in $C$.

The other direction comes from definition of a metric.

oliverjones
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