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I understand that if A is open in the metric space (M,d), then for any $G \subset A$ we have that G is open in A iff G is open in M. But if we replace the word "open" with "closed", I don't believe it's true. I'm trying to find a counterexample using a subset of $\mathbb{R}$, to no avail.

elmo
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  • This seems to have been answered: https://math.stackexchange.com/questions/313991/is-it-true-that-a-subset-that-is-closed-in-a-closed-subspace-of-a-topological-sp – Chris Eagle Aug 12 '19 at 21:58
  • well let $G = A$. $A$ is both open and closed in $A$ but it is only open in $M$. – fleablood Aug 12 '19 at 22:04
  • Or for that matter if $A = (0,2)\subset \mathbb R$ and $G = [1,2)$ is closed in $A$. If $G$ is a set that contains all it's limit points that are in $A$ but none of its limit points that aren't in $A$... – fleablood Aug 12 '19 at 22:07
  • The comments above, and the answer, assume that you still want $A$ to be open, and are just asking about the closedness of $G$. The link I posted is for the case where $A$ is also taken to be closed. Which instances of "open" were supposed to be replaced by "closed"? – Chris Eagle Aug 12 '19 at 22:33
  • Sorry I meant to replace all instances of "open". So A is presumed closed too. – elmo Aug 13 '19 at 07:21
  • @ChrisEagle That link proves the forward implication. But I think the backwards implication is also true. Let $A$ as well as $G \subset A$ be closed in M. Since $G = G \cap A$, G is closed in A. – elmo Aug 13 '19 at 07:31
  • If $G\subset A$ then every limit point of $G$ is a limit point of $A$. If $A$ is closed then every limit point of $G$ is a point in $A$. So $G$ is closed in $A$ if and only if every limit point of $G$ that is in $A$ is in $G$ if and only if every limit point of $G$ that is in $M$ is in $G$ if and only if $G$ is closed in $M$. – fleablood Aug 13 '19 at 14:50

2 Answers2

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The definition of $G$ being closed in $A$ means that $G$ contains all its limit points (that are in $A$), whereas the definition of $G$ being closed in $M$ means that $G$ contains all it limit points (in $M$).

So to find a counter-example we need a set $G$ that has limit points that are not in $A$ but of all the limit points that are in $A$ are in $G$.

So let's take a case where $G$ is not closed. Let's say $G= (0,1)$. It has limit points $0, 1$ and all the points $(0,1)$. $0$ and $1$ are not in $G$ so if $0,1$ are not in $A$ we will be fine. The points $x: 0 < x < 1$ are all in $G$ so $(0,1)\subset A$ but $0,1 \not \in A$ will be good.

SO if $A = (0,1)$ then $G=(0,1) \subset A$ is closed in $A$.

For a less trivial example so $A = (0,1) \cup (2,3)$. $G=(0,1)$ is closed in $A$.

For a more subtle case. If $G = [0,1)$ then $G$ has all its limit points except $1$ in $G$. If $1\not \in A$ and $G\subset A$ and $A $ is open this will work. Let $A = \mathbb R\setminus \{1\}$. Then $G$ is closed in $A$ but not in $\mathbb R$.

fleablood
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If $A$ is closed in $M$ then $B \subseteq A$ is closed in $A$ iff $B$ is closed in $M$. This is immediate from the definition of the subspace topology, and holds in all spaces, not just metric ones.

Henno Brandsma
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