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I need to prove that:

$$d(a,X) = 0 \iff X\cap U \neq \emptyset$$ for all open set $U$ that contains $a$

My idea is that if $d(a,X) = 0$, then there is a point $b\in X$ such that $d(a,b)=0$. In some way, I should be able to construct a ball that contains $a$ and $b$. Remember that $b\in X$ so the intersection should not be empty.

Any ideas on how to fill the gap I left in my proof?

Andrés E. Caicedo
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Guerlando OCs
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  • Should that be "for every open set $U$ containing $a$" instead of "for some $U$ containing $a$"? Otherwise the claim you try to prove is not even correct. – Frank Lu Apr 11 '16 at 02:17
  • @FrankLu yes, gonna correct that – Guerlando OCs Apr 11 '16 at 02:23
  • $d(a.X)$ is, by def'n, $\inf {d(x,y):y\in X}.$ And $ d(a,X)=0$ does NOT imply $\exists b\in X;(d(a,b)=0).$ For example with the usual metric on the reals and $X=(0,1]$ we have $d(0,X)=0.$ – DanielWainfleet Apr 11 '16 at 06:13

2 Answers2

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"$\Rightarrow$" Suppose $d(a,X)=0$, then for any $\epsilon>0$, there exists $b_\epsilon\in X$ such that $d(a,b_\epsilon)<\epsilon$. Let $B(a,\epsilon)$ denote the open ball centered at $a$ with radius $\epsilon$, then $B(a,\epsilon)\cap X\neq \emptyset$. As $\epsilon>0$ is arbitrary, we conclude that every open ball centered at $a$ intersects $X$. Thus this direction is proved.

"$\Leftarrow$": Suppose $X\cap U\neq\emptyset$ for every open set containing $a$. Then we consider two cases:

  • If $a\in X$, then there is nothing to prove.
  • If $a\notin X$, note that for each $\epsilon>0$, we have $X\cap (B(a,\epsilon)\setminus\{a\})\neq\emptyset$, so there exists $b_\epsilon\in X\cap(B(a,\epsilon)\setminus\{a\})$, in particular $d(a,b_\epsilon)<\epsilon$. By letting $\epsilon\to 0$, the claim follows.
Frank Lu
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I use the following equivalence: $a$ belongs to $\operatorname{Cl} X$ iff for all nbd's $N$ of $a$, $N\cap X\not=\varnothing$

Since $0=d(a,X)=\inf\{d(a,x):x\in X\}$, we can choose $x_n$ in $X$ such that $d(a,x_n)<n^{-1}$ for all $n$. So, there is a sequence $\{x_n\}\subset X$ such that $x_n \to a$, that is, $a$ belongs to the closure of $X$.

Conversely, if $a$ belongs to $\operatorname{Cl} X$, we can find a sequence $\{x_n\}$ of elements in $X$ such that $x_n\to a$ (only consider the nbd's $\{B(a,n^{-1})\}$ and use that $B(a,n^{-1})\cap X\not= \varnothing$) and this implies $d(a,X)=0$.

Jose Antonio
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