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In a metric space $X$, or just $R^K$, $E \subset X$ is a closed set if every limit point of $E$ is a point of $E$. Is it possible to use only this fact, or definition, to prove that a finite point set is closed?

I know we can prove that because the complement of a finite point set is open. However, since a finite point set has no limit points as every neighborhood of a limit point should contain infinitely many points of $E$, I cannot figure out how to use this concept of limit points to prove that a finite point set is closed.

Thank you in advance!

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    $\begingroup$ If $E$ has no limit points, then all of them are vacuously elements of $E.$ $\endgroup$ – saulspatz Oct 20 '18 at 17:43
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Recall that a closed set (in a metric space) is a set which contains all of its limit points (note that this is only one of many commonly use equivalent definitions). Since a finite set has no limit points (proved below), we can conclude that it contains all of its limit points (since there are none).

Therefore, any finite subset of a metric space is closed.

Proposition: If a set $X$ in a metric space has a limit point $x$, then $X$ is infinite.

Proof. Consider the ball of radius $r=1$ centred at $x$. By our assumption, this ball contains a point in $X$ that is distinct from $x$. Call this point $x_{1}$. Now, consider the ball of radius $r=d(x,x_{1})$ centred at $x$. Again, this ball contains a point in $X$ that is distinct from $x$. Moreover, this new point cannot equal to $x_{1}$ due to our choice of radius. Call this new point $x_{2}$. Continuing this way, we obtain a sequence of points in $X$ which are all distinct from each other. Therefore, $X$ is infinite.

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  • $\begingroup$ I don't get the meaning of being "trivially" closed. Does that mean that because a finite point set is not open, it is closed...? $\endgroup$ – Hunnam Oct 20 '18 at 19:24
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    $\begingroup$ Trivially just means "easily". I added more detail. $\endgroup$ – parsiad Oct 21 '18 at 4:27

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