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My question is related to proving that any subset $D \subset X$, where $(X,d)$ is a metric space with $d$ being the discrete metric, is both open and closed.

I've read some suggestions to a solution, such as the accepted answer from this question: Show that in a discrete metric space, every subset is both open and closed., which basically amounts to proving that $D$ contains a ball of radius 1 around each of its points.

I understand that if we choose a ball of radius 1 for any $x_0 \in D$, the ball will only contain its centerpoint $x_0$.

But what if we chose a ball with radius larger than 1, say $r$? Wouldn't the ball consist of a sphere with points $\{x \in X: d(x,x_0) = 1\}$, and the centerpoint $x_0$? I.e. choosing $r > 1$ gives that each point in $D$ is surrounded with a sphere of radius 1.

Does this in any way contradict the result of the proof; "... any $D$ is both open and closed"?

My intuition is that the sphere argument is invalid somehow, but I don't understand why. I need help understanding why choosing a ball of radius 1 is sufficient to make the above conclusion.

Community
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harisf
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  • A ball with radius larger that $1$ with the discrete metric is the entire space. – Tim Raczkowski Jul 22 '15 at 15:18
  • Oh. But the definition of the discrete metric states that the distance is equal to 1 for two dissimilar elements. Can you elaborate? – harisf Jul 22 '15 at 15:23
  • Let $a\in X$. Clearly, $a\in B(a,r)$ for $r>1$ and if $b\in X$ with $b\ne a$, then $d(a,b)=1<r$. So, $b\in B(a,r)$. Hence, $B(a,r)=X$. – Tim Raczkowski Jul 22 '15 at 15:31
  • Thanks, @TimRaczkowski. I think I get why the ball around a point is the whole space, if $r > 1$. – harisf Jul 22 '15 at 16:01

2 Answers2

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The open ball $\{y\in X : d(x,y) < \frac{1}{2}\} = \{x\}$

Hence every single point set is open.

Can you proceed?

Brian
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  • In the case for a closed subset we can proceed with closed balls? – Masacroso Jul 22 '15 at 15:17
  • @syuizen The question is, what if the ball is chosen such that the $r>1$. I get that $r \leq 1$ gives a set containing only the centerpoint. – harisf Jul 22 '15 at 15:19
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    @Masacroso Yes. If you try to show the complement set is open , then you don't need to think of closed ball.Anyway , as you like. – Brian Jul 22 '15 at 15:20
  • @harisf if $r>1$ then the ball is the whole space $X$. – Brian Jul 22 '15 at 15:23
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So, choosing $r> 1$ when considering a ball around a point $x \in D$, gives the whole space $X$ on which the discrete metric is defined. That is why the proof restricts to $r \leq 1$.

There is no such thing as a sphere consisting of points on $r = 1$ in this case.

harisf
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