Prove that the intersection of subspaces is compact and closed

[docnet]

Homework Statement

If $X$ is a topological space and $S_i \subset X$ is an arbitrary collection of closed subspaces, at least one of which is compact, then $\bigcap_i S_i$ is also closed and compact.

Relevant Equations

(o.o)_)~

Given that one of the $S_i$ (let's name it $S_{compact}$), is compact. Assume there is an open cover $\mathcal V$ of $S_{compact}$. By definition of a compact subspace, there is a subcover $\mathcal U$ with $n<\infty$ open sets. Notice that $\forall x\in (\bigcap_i S_i)$, $x\in S_i$, which implies that the intersection of all the $S_i$'s is contained in anyone of the $S_i$'s. Since the intersection of the $S_i$'s is contained in $S_{compact}$, the subcover for $S_{compact}$ is also a subcover of the intersection of the $S_i$'s. Also, given that every $S_i$ is closed, their arbitrary intersection is also closed. It follows that $\bigcap S_i$ is closed and compact.

 

[jim mcnamara]

And your question is?

 

[fresh_42]

Homework Statement:: If $X$ is a topological space and $S_i \subset X$ is an arbitrary collection of closed subspaces, at least one of which is compact, then $\bigcap_i S_i$ is also closed and compact.
Relevant Equations:: (o.o)_)~

Given that one of the $S_i$ (let's name it $S_{compact}$), is compact. Assume there is an open cover $\mathcal V$ of $S_{compact}$. By definition of a compact subspace, there is a subcover $\mathcal U$ with $n<\infty$ open sets. Notice that $\forall x\in (\bigcap_i S_i)$, $x\in S_i$, which implies that the intersection of all the $S_i$'s is contained in anyone of the $S_i$'s. Since the intersection of the $S_i$'s is contained in $S_{compact}$, the subcover for $S_{compact}$ is also a subcover of the intersection of the $S_i$'s. Also, given that every $S_i$ is closed, their arbitrary intersection is also closed. It follows that $\bigcap S_i$ is closed and compact.

Your proof about the closure is correct. Arbitrary intersections of closed sets are closed, because arbitrary unions of open sets are open (standard axiom to define a topology):
\begin{align*}
X-\cap S_i =\underbrace{ \cup \underbrace{(X - S_i)}_{\text{open}}}_{\text{open}} \Longrightarrow \cap S_i \text{ closed}
\end{align*}

The proof of compactness is not complete. It starts with any given open cover $\mathcal{V}$ of $\cap S_i$. Say $S_1$ is compact. How do we get an open cover for $S_1$ to which we can apply the compactness condition? We do not have an open cover of $S_1$, yet! Can you give us one?

Hint: I treated the closeness part first on purpose!

 

[docnet]

Your proof about the closure is correct. Arbitrary intersections of closed sets are closed, because arbitrary unions of open sets are open:
\begin{align*}
X-\cap S_i =\underbrace{\cup \underbrace{X-S_i}_{\text{open}}_{\text{open} \Longrightarrow \cap S_i \text{ closed}}}
\end{align*}

The proof of compactness is not complete. It starts with any given open cover $\mathcal{V}$ of $\cap S_i$. Say $S_1$ is compact. How do we get an open cover for $S_1$ to which we can apply the compactness condition? We do not have an open cover of $S_1$, yet! Can you give us one?

hmm I could first construct an open cover $\mathcal W$ of $S_1\backslash (\cap S_i)$ and use that to construct an open cover of $S_1$ by writing $\mathcal W \cup \mathcal V$. Then apply the compactness condition on $S_1$ to get a subcover $\mathcal U$ for any $\mathcal W \cup \mathcal V$. Then the subcover of $\cap S_i$ could be constructed as $\mathcal U \cap (\cap S_i)$. Is this kind of close to what you were thinking??

 

[fresh_42]

hmm I could first construct an open cover $\mathcal W$ of $S_1\backslash (\cap S_i)$ and use that to construct an open cover of $S_1$ by writing $\mathcal W \cup \mathcal V$. Then apply the compactness condition on $S_1$ to get a subcover $\mathcal U$ for any $\mathcal W \cup \mathcal V$. Then the subcover of $\cap S_i$ could be constructed as $\mathcal U \cap (\cap S_i)$. Is this kind of close to what you were thinking??

We have a collection $\mathcal{V}$ of open sets which cover $\cap S_i$. And we have an open set $X-\cap S_i$. What does $(X-\cap S_i) \cup (\cup_{U\in \mathcal{V}} U)$ cover?

 

[docnet]

We have a collection $\mathcal{V}$ of open sets which cover $\cap S_i$. And we have an open set $X-\cap S_i$. What does $(X-\cap S_i) \cup (\cup_{U\in \mathcal{V}} U)$ cover?

oopsie! I now see that it covers all of $X$. If I'm not wrong again, $(X-\cap S_i) \cup (\cup_{U\in \mathcal{V}} U)$ is a subcover for X, which implies $(\cup_{U\in \mathcal{V}})$ is a subcover of the union of the $S_i$'s, which concludes the proof?

 

[fresh_42]

oopsie! I now see that it covers all of $X$. If I'm not wrong again, $(X-\cap S_i) \cup (\cup_{U\in \mathcal{V}} U)$ is a subcover for X, which implies $(\cup_{U\in \mathcal{V}})$ is a subcover of the union of the $S_i$'s, which concludes the proof?

$\mathcal{W}:=\{\mathcal{V},X-\cap S_i\}$ is an open cover of $X$, not a subcover.
But as it covers all, it especially covers the compact set $S_1.$ This means that there is a finite subcover $\mathcal{K}\subseteq \mathcal{W}$ of $S_1$. Now $\cap S_i \subseteq S_1$ so $\mathcal{K}$ is an open and finite cover of $\cap S_i$, too.

Finally, you must show that $\mathcal{K}$ is a subcover of $\mathcal{V}.$ We only have $\mathcal{K}\subseteq \mathcal{W}$ from the compactness of $S_1$, but we need $\mathcal{K}\subseteq \mathcal{V}$.

The structure to prove compactness of a set $A$ is always: Given any open cover $\mathcal{V}$ of $A$, show that there are finitely many open sets $U_1,\ldots,U_n \in \mathcal{V}$ such that $A\subseteq \cup_{i=1}^n\;U_i.$

Don't let $n$ go undefined as in your first post. It is the number of sets in your given cover that already covers the given set, the order (= number of elements) of the subcover.

 

[docnet]

$\mathcal{W}:=\{\mathcal{V},X-\cap S_i\}$ is an open cover of $X$, not a subcover.
But as it covers all, it especially covers the compact set $S_1.$ This means that there is a finite subcover $\mathcal{K}\subseteq \mathcal{W}$ of $S_1$. Now $\cap S_i \subseteq S_1$ so $\mathcal{K}$ is an open and finite cover of $\cap S_i$, too.

Finally, you must show that $\mathcal{K}$ is a subcover of $\mathcal{V}.$ We only have $\mathcal{K}\subseteq \mathcal{W}$ from the compactness of $S_1$, but we need $\mathcal{K}\subseteq \mathcal{V}$.

The structure to prove compactness of a set $A$ is always: Given any open cover $\mathcal{V}$ of $A$, show that there are finitely many open sets $U_1,\ldots,U_n \in \mathcal{V}$ such that $A\subseteq \cup_{i=1}^n\;U_i.$

Don't let $n$ go undefined as in your first post. It is the number of sets in your given cover that already covers the given set, the order (= number of elements) of the subcover.

A Re-attempt for a coherent a proof:

Given that one of the $S_i$'s (let's name it $S_{1}$) is compact, assume there is an open cover $\mathcal V$ of $S_{1}$. By definition of a compact subspace, there is a subcover $\mathcal U$ with $n<\infty$ open sets, where $n$ is the order of the subcover of $S_1$. Notice that $\forall x\in (\bigcap_i S_i)$, $x\in S_i$ which implies that the intersection of all the $S_i$ is contained in any of the $S_i$'s. Since the intersection of all the $S_i$'s is contained in $S_{1}$, there is a subcover $\mathcal K$ over $\cap S_i$ that is contained in $\mathcal U$. The open subsets in the subcover $\mathcal K$ are the $K$'s , which are the intersection of each $U\in\mathcal U$ and $\cap S_i$. In other words $K \in \mathcal K = \{(\cap S_i) \cap (U\in\mathcal U)\}$. Since the number of open sets in $\mathcal U$ is $n$, the number of those open sets that intersect with $\cap S_i$ is $\leq n$, which is also a finite number. Also, given that every $S_i$ is closed, their arbitrary intersection is also closed. It follows that $\bigcap S_i$ is closed and compact.

I hope I addressed everything @fresh_42, and as always thank you for your help!

 

[fresh_42]

$(\cap S_i) \cap U$ is not open.

We already have had almost the entire proof (steps 1-6) and only step 7 was missing:

1.) $X-\cap S_i$ is open.
2.) $\cap S_i$ is closed.
3.) Let $\mathcal{V}$ be an open cover of $\cap S_i$.
(We have to find a finite subcover of $\mathcal{V}$ that already covers $\cap S_i$.)
4.) $\mathcal{W}:=\{\mathcal{V},X-\cap S_i\}$ is an open cover of the compact set $S_1$.
5.) There are finitely many sets $U_1,\ldots,U_n \in \mathcal{W}$ which cover $S_1$, because $S_1$ is compact.
6.) $U_1,\ldots,U_n \in \mathcal{W}$ cover $\cap S_i$, since $\cap S_i\subseteq S_1.$ Say $\mathcal{K}:=\{U_1,\ldots,U_n\}.$
7.) If all $U_k\in \mathcal{V}$ then we are done.
Assume w.l.o.g. that $U_1\not\in \mathcal{V}$. Then $U_1=X-\cap S_i$ because all other sets are in $\mathcal{V}$. But $(\cap S_i)\cap U_1=\emptyset$. Hence we can drop $U_1$ from the cover $\mathcal{K}$ and still have a cover of $\cap S_i$. But these sets are all in $\mathcal{V},$ i.e. we have found finitely many sets in $\mathcal{V}$ whose union contain $\cap S_i$, i.e. a finite subcover of $\mathcal{V},$ i.e. $\cap S_i$ is compact.

A cover of a set $A$ is a set of sets $U_\iota$, whose union contains $A$.
An open cover of a set $A$ is a set of open sets $U_\iota$, whose union contains $A$.
A subcover of a cover of $A$ is a set of sets $U_i$ from a given cover of $A$, whose union contains $A$. It is a subset of the set of sets that cover $A$.
A finite subcover is a subcover with only finitely many sets.

Try to go step by step through a proof, beginning with what you have ($S_i$ closed, $S_1$ compact) and ending with what you need ($\cap S_i$ closed, $\cap S_i$ compact).

You remind me of a real-life tutorial I once had. She guessed solutions and looked into my face, trying to figure out how far from the real solution she was. Only problem is, that you cannot see my face. Uhmm, I think you can't. Uhmm, have you hacked my camera?

 

[docnet]

@fresh_42 Oh my gosh! thank you so much for explaining step by step how to solve this problem!
I can't write an elegant proof like that on my own yet. I think I am starting to see why great proofs are nice!

 

[PeroK]

@fresh_42 Oh my gosh! thank you so much for explaining step by step how to solve this problem!
I can't write an elegant proof like that on my own yet. I think I am starting to see why great proofs are nice!

Can you prove that the intersection of two compact sets is compact? I suggest that as an exercise.

 

[docnet]

Can you prove that the intersection of two compact sets is compact? I suggest that as an exercise.

Then I'll closely follow @fresh_42's steps (because I'm prone to getting lost).

1.) $X- A \cap B$ is open.
2.) $A \cap B$ is closed.
3.) Let $\mathcal{V}$ be an open cover of $A \cap B$.
(We have to find a finite subcover of $\mathcal{V}$ that already covers $A \cap B$.)
4.) $\mathcal{W}:=\{\mathcal{V},X-A \cap B\}$ is an open cover of the compact set $A$.
5.) There are finitely many sets $U_1,\ldots,U_n \in \mathcal{W}$ which cover $A$, because $A$ is compact.
6.) $U_1,\ldots,U_n \in \mathcal{W}$ cover $A \cap B$, since $(A\cap B)\subseteq A.$ Say $\mathcal{K}:=\{U_1,\ldots,U_n\}.$
7.) If all $U_k\in \mathcal{V}$ then we are done.
Assume w.l.o.g. that $U_1\not\in \mathcal{V}$. Then $U_1=X-(A\cap B)$ because all other sets are in $\mathcal{V}$. But $(A\cap B)\cap U_1=\emptyset$. Hence we can drop $U_1$ from the cover $\mathcal{K}$ and still have a cover of $A\cap B$. But these sets are all in $\mathcal{V},$ i.e. we have found finitely many sets in $\mathcal{V}$ whose union contain $A\cap B$, i.e. a finite subcover of $\mathcal{V},$ i.e. $A\cap B$ is compact.

 

[PeroK]

Then I'll closely follow @fresh_42's steps (because I'm prone to getting lost).

1.) $X- A \cap B$ is open.
2.) $A \cap B$ is closed.
3.) Let $\mathcal{V}$ be an open cover of $A \cap B$.
(We have to find a finite subcover of $\mathcal{V}$ that already covers $A \cap B$.)
4.) $\mathcal{W}:=\{\mathcal{V},X-A \cap B\}$ is an open cover of the compact set $A$.
5.) There are finitely many sets $U_1,\ldots,U_n \in \mathcal{W}$ which cover $A$, because $A$ is compact.
6.) $U_1,\ldots,U_n \in \mathcal{W}$ cover $A \cap B$, since $(A\cap B)\subseteq A.$ Say $\mathcal{K}:=\{U_1,\ldots,U_n\}.$
7.) If all $U_k\in \mathcal{V}$ then we are done.
Assume w.l.o.g. that $U_1\not\in \mathcal{V}$. Then $U_1=X-(A\cap B)$ because all other sets are in $\mathcal{V}$. But $(A\cap B)\cap U_1=\emptyset$. Hence we can drop $U_1$ from the cover $\mathcal{K}$ and still have a cover of $A\cap B$. But these sets are all in $\mathcal{V},$ i.e. we have found finitely many sets in $\mathcal{V}$ whose union contain $A\cap B$, i.e. a finite subcover of $\mathcal{V},$ i.e. $A\cap B$ is compact.

I was confident you would find a proof, despite the fact that in general the proposition in false! If you assume that compact sets must be closed (which is true in $\mathbb R^n$ and more generally in any Hausdorff space), then the intersection of compact sets is compact. But, in general, a compact set need not be closed. There's a counterexample here:

https://math.stackexchange.com/ques...n-of-finite-number-of-compact-sets-is-compact