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I have seen this proved always by taking the complement. Thus I am wondering if the following argument is correct.

Here $(X,d)$ is a metric space.

Question: Show that the set $\{y\in X : d(x,y) \leq r\}$, called a closed ball, is a closed set.

My Proof: Let us denote $\bar{B}(x;r) = \{y\in X : d(x,y) \leq r\}$. For this to be closed the following must hold: if $z \in \bar{B}(x;r)$ then for every $s>0$, $B(z;s) \cap B(x;r) \neq \emptyset$. This is true since $d(x,z) \leq r$, thus the two balls always overlap.

geguze
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2 Answers2

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Your argument also shows that the open ball is closed. Let $B = \{y \in X : d(x,y) < r\}$, an arbitrary open ball. For all $z \in B$ and all $s > 0$, $B(z; s) \cap B \neq \varnothing$. By your argument, $B$ is closed.

Note that the nonempty intersection of the two balls holds for every point in the closure of a set. That is, for $U \subset (X,d)$, for all $z$ in the closure of $U$, a ball around $z$ intersects $U$. This is true even if $U$ is open.

Your intersecting balls criterion shows that $z$ is a limit point. Every point in the closure of a set is a limit point of the set. You would need to show that every limit point of $\overline{B}(x;r)$ is actually contained in $\overline{B}(x;r)$.

Eric Towers
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  • Thanks Eric. I definitely see the point you are making. I am not sure I understand how to use the suggestion you make in the last paragraph. – geguze Apr 01 '18 at 22:08
  • So what I proved is only that $z$ is adherent to $B(x;r)$, but clearly this is not enough to show that $B(x;r)$ is closed. Probably what you are suggesting, regarding limit points, is something I have not yet reached in the book, as I don't know of a characterization of closeness based on limit points, yet ... – geguze Apr 01 '18 at 22:22
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In a general metric space, it can happen that your $B(z;s)$ and $B(x,r)$ do not overlap. Indeed, consider any space $X$ with the discrete metric $$d(x,y)=\begin{cases}0&\text{if }x=y\\1&\text{otherwise}\end{cases}$$ Then $\overline B(x; 1) = X$ and $B(x;1)=\{x\}$ for all $x\in X$. In particular, if $X$ has at least two distinct elements $x,z$, then $z\in\overline B(x;1)$, but $B(x; 1)$ and $B(z;1)$ do not overlap.

Are you sure that your criterion for closedness is correct?

Hagen von Eitzen
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