$M$ is metric space. $A\subset M, F$ is closed in $M$, prove $A\cap F$ is closed in $A$

Question

$M$ is metric space, $A\subset M, F$ is closed in $M$, I'm asked to prove $A\cap F$ is closed in $A$.

First of all, what does "closed in $A$ mean"?

Does it mean that $A$ is now the metric space being considered?

If this is the case and a metric space is both open and closed, why can't I just say $A\cap F$ is closed because it is an intersection of two closed sets?

Answer 1

When $A$ is a subset of a metric space $(X,d)$ we consider it to be a metric space in its own right $(A, d_A)$, where $d_A$ is just the restriction of $d$ from $X \times X$ to $A \times A$: we only compute the distances between points of $A$, as this is the new "universe".

Any metric space defines a topology on its underlying set, in a way you probably know (or should). So $(A, d_A)$ also defines a topolgoy on $A$, in which we can talk about closed sets and open sets (and compact sets, connected sets, continuity etc.)

So now $F$ is closed in $X$, then $F \cap A$ is a subset of $A$.

We can use the following characterisation of closed sets in any metric space $Y$:

$H\subseteq Y$ is closed in $Y$ iff for every sequence $(y_n)$ in $H$ such that $y_n \to y$ in $Y$ (the universe set) we can conclude that $y \in H$.

Now let $(a_n)$ be a sequence in $A \cap F$ such that in $A$ we have $a_n \to a \in A$. Because converge of a sequence is purely defined in terms of the metric ($\forall \epsilon>0: \exists N: \forall n \ge N: d_A(a_n, a) < \epsilon$ for the converge of $a_n$ to $a$) and as $d_A(a,a')=d(a,a')$ for all $a,a' \in A$ (that's what being a restriction means) $a_n \to a$ in $(X,d)$ as well. As all $a_n \in F$ ($A \cap F \subseteq F$) by the same characterisation we conclude that $a \in F$ as $F$ is closed in $X$. Hence $a \in A \cap F$ and we have shown that $A \cap F$ is closed in $A$.

Answer 2

Openness and closedness only make sense to talk about when you've fixed a metric space to work in. You're told that $F$ is closed as a subspace of $M$. Restricting the metric from $M$, you can talk about $A$ as a metric space in its own right, and then yes you're right $A$ is closed in itself (trivially). What's you're asked to show is that $F \cap A$ is closed as a subspace of $A$. This is not immediate, you need to go back to whatever definition you're using for closedness.

Is your definition of closed that it contains the limits of all sequences? Then you need to confirm that if a sequence of points in $F \cap A$ has a limit in $A$, that limit is in fact in $F \cap A$.

Is your definition of closed that the complement is open? Then you need to show that given $a \in F - A$, there exists a ball $B$ (in the metric on $A$) around $a$ such that $B \cap (F \cap A) = \emptyset$.

Answer 3

If $(M,d)$ is a metric space and $A\subseteq M$ then - if $d'$ denotes the restriction of $d$ on $A\times A$ - also $(A,d')$ is a metric space.

For a set $F\subseteq M$ that is closed in metric space $(M,d)$ you must prove that the set $A\cap F$ is closed in metric space $(A,d')$.

In this context $F$ is a closed set of $M$ and $A$ is (as you reason) a closed set of $A$, but it does not speak for itself that you can conclude from this that $A\cap F$ is a closed set of $A$ unless you practicize the definition of subspace topology. This makes me I suspect that you must give a proof here specifically for metric spaces and reveals consistency with the topological concept "subspace topology".

A correct way is proving that the set $A\cap F^{\complement}=A\setminus F$ is open in metric space $(A,d')$.

Let $x\in A\setminus F$.

Then for $\epsilon>0$ small enough we have $\{y\in M\mid d(y,x)<\epsilon\}\subseteq F^{\complement}:=X\setminus F$ so that consequently $\{y\in A\mid d'(y,x)<\epsilon\}\subseteq A\cap F^{\complement}$.

Apparantly for every $x\in A\setminus F$ we can find an $\epsilon>0$ such that ball $B(x,\epsilon)$ in metric space $(A,d')$ is a subset of $A\setminus F$.

This proves that $A\setminus F$ is open in $(A,d')$.