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Is every finite subset of a metric space closed ? Yes/No

My attempt : i think No

Consider $X = \{0, 1\}$ with the indiscrete metrics i.e. the only two open sets are $\emptyset$ and $X = \{0, 1\}$ itself.

the singleton subset $\{0\}$ of $X$ is not closed because its complement $X \setminus \{0\} = \{1\}$ is not one of the two open sets listed above and therefore not open.

Is its true ?

Community
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jasmine
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    There is no such thing as an indiscrete metric (unless, of course, you are considering it on the topological spaces ${0}$ or $\emptyset$). –  Dec 06 '19 at 14:05
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    You seem to be mixing up metric and topology. – Arnaud Mortier Dec 06 '19 at 14:07
  • [email protected]. but that is definition of indiscrete metrics – jasmine Dec 06 '19 at 14:07
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    That space isn't Hausdorff, so it can't be a metric space. – nbritten Dec 06 '19 at 14:08
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    This is an indiscrete topology, not indiscrete metric. Such a topological space is not metrizable. And by the way, the statement is actually true. Try to prove that a singleton is closed in a metric space, then use the fact that a finite union of closed sets is closed. – Mark Dec 06 '19 at 14:09
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    but that is definition of indiscrete metrics --- An explicit citation that can be examined by others would go a long way towards understanding the nature of your confusion. – Dave L. Renfro Dec 06 '19 at 15:59

2 Answers2

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Suppose $\{p_1,...,p_n\} \subset X$ and $x \notin F$. Let $\epsilon = \min_kd(x,p_k)$. Then $B(x,\epsilon) \cap F = \emptyset$ and so $F^c$ is open.

copper.hat
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Let $(X, d)$ be a metric space and $x\in X$ be given. For each $x\neq y\in X$ define $O_y = B(y, d(x, y))$. Note that each $O_y$ is open, contains $y$, and does not contain $x$. Then $O = \bigcup_{y\neq x} O_y$ is open and $O = X\setminus \{x\}$. So $\{x\}$ is closed as the complement of an open set.

Now let $x_1, \ldots, x_n\in X$ be given. From the previous paragraph each $\{x_i\}$ is closed, so $\bigcup_{i = 1}^n \{x_i\} = \{x_1, \ldots, x_n\}$ is closed as the finite union of closed sets.

nbritten
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