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$M$ 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?

Andrews
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Heuristics
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  • Do you know concept of metric sub space – neelkanth Feb 18 '19 at 17:47
  • @neelkanth I don't think its mentioned in my litterature, and I don't know it. – Heuristics Feb 18 '19 at 17:53
  • You have to think about where your set $F$ is closed. You know that it is closed in $M$. This does not automatically mean that it is closed in $A$, as there is no a priori reason to assume that the topology in $A$ is "compatible" with the topology in $M$. The goal of the exercise is to show that, indeed, the topologies are compatible. I would give a more in depth answer, but there are some key details missing from your question. For example, how, exactly, are you defining "closed sets"? – Xander Henderson Feb 18 '19 at 17:53
  • Because F might not be closed in A. In fact, F be out of A, in A, or just have some part of intersection with A. – caden Hong Feb 18 '19 at 17:53
  • Where did you see this exercise? In a course about topology or analysis? Because in topology, that's basically the definition of subspace topology. – stressed out Feb 18 '19 at 17:56
  • Ack... I linked to the wrong question. This is, I think, a duplicate of https://math.stackexchange.com/questions/1940995/show-that-a-subset-f-of-y-is-closed-in-y-if-and-only-if-f-y-cap-h-for-s . – Xander Henderson Feb 18 '19 at 18:31
  • Or this: https://math.stackexchange.com/questions/1626251/show-that-f-subset-y-is-closed-in-y-iff-f-y-cap-h-where-h-subset-x-is – Xander Henderson Feb 18 '19 at 18:32
  • Applying the definition of subspace topology $A\cap F$ is closed in $A$ by definition. You are asked to prove that this also works if $A$ is looked at as a metric subspace of $M$ equipped with restricted distance. – drhab Feb 18 '19 at 18:42

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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 $B \cap (F \cap A) = \emptyset$.

nkm
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    This is not really an answer. Instead, it is a request for clarification from the asker, and should be posted as a comment. – Xander Henderson Feb 18 '19 at 18:00
  • I answered both of the actual questions posed, about what it means for a set to be closed in $A$, and about why he can't just cite the intersection of two closed sets. I didn't finish the homework problem, but that wasn't actually requested. Also people post "answers" like this all the time! I don't mean to cause trouble, but I certainly think I was adding value. – nkm Feb 18 '19 at 18:17
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    I was not suggesting that you answer the homework question. What I was suggesting is that the question is unclear, and that your answer would better be left as a comment seeking clarification. The fact that others put comments in the answer box does not indicate that this is good practice. – Xander Henderson Feb 18 '19 at 18:19
  • The question was not unclear. He asked two specific questions about the subspace topology, which I answered. – nkm Feb 18 '19 at 18:20
  • Clearly I'm losing this battle. I was neither critiquing nor requesting clarification. I answered all questions (identified with question marks) in the original post. I frankly find this culture of policing the line between comment and answer ridiculous. – nkm Feb 18 '19 at 18:50
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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 $(A,d')$.

drhab
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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 $A$.

Henno Brandsma
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