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I have 2 contradictory solutions to the follower problem:

The problem:

Let $X$ be an infinite set. For $p \in X$ and $q\in X$ define $$ d(x,y)= \begin{cases} 0\qquad&\text{if and only if $x=y$}\\\ 1&\text{otherwise} \end{cases} $$ Which subsets are closed and which are open?

My "solution":

Every point is isolated: Consider balls of radius $1/2$ around each x. Which is to say no point is a limit point in $X$ (a limit point will have other points of $X$ in the neighborhood). These points, which are not limit points, will be contained in any subset of $X$. Thus no subset of $X$ will contain limit points. Thus no subset of $X$ will be closed.

But:

Balls of radius $1/2$ centered at $x \in X$ are open because the topology of a metric space has all open balls defined to be open. The complement of this ball is closed by proof in Rudin. It is also non-empty (containing all the other points that are not $x\in X$). Thus the open sets are unions of open balls containing points and their complements are the closed sets.

I produced no closed sets in the first argument and a bunch in the second -- what happened?? After reviewing the "similar questions", I think the second explanation is correct. I don't see why the first one is wrong though.

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    $\begingroup$ a closed set does not have to have limit points. you proved every subset is open, and so every subset is closed also. $\endgroup$ – Forever Mozart Feb 22 '18 at 2:54
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    $\begingroup$ A set is closed if and only if it contains all of its limit points. If the set has no limit points in the first place then this statement is vacuously true. So your conclusion "Thus no subset of $X$ will contain limit points. Thus no subset of $X$ will be closed" is not true. $\endgroup$ – wgrenard Feb 22 '18 at 2:58
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    $\begingroup$ The over reliance on limit points is a blunder, an inferior approach to closed sets. $\endgroup$ – William Elliot Feb 22 '18 at 3:04
  • $\begingroup$ @wgrenard So the subsets of the integers under $R^1$ metric are closed vacuously? $\endgroup$ – yoshi Feb 22 '18 at 3:09
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    $\begingroup$ @yoshi The second argument looks correct, but I don't think it's a good answer for the problem. The correct answer may be "All subsets in $X$ are both open and closed", in other words, the topology of $X$ is discrete. $\endgroup$ – ChoF Feb 22 '18 at 3:29

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