Counterexample: relatively closed subset in A but not in the whole metric space

Question

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.

Answer 1

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 $\mathbb R$.

Answer 2

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.