A proof that a union of compact spaces is compact

[docnet]

Homework Statement

prove that a union of compact spaces is compact

Relevant Equations

o.o

Prove that if $X$ is a topological space, and $S_i \subset X$ is a finite collection of compact subspaces, then their union $S_1 \cup \cdots \cup S_n$ is also compact.

$S_i \subset X$ is compact $\therefore \forall S_i, \exists$ a finite open cover $\mathcal J_i=\{U_j\}_{j\in J\subset K} \subset \{U_k\}_{k\in K}=\mathcal K_i$, where $\mathcal K_i$ is an open cover of $S_i$. I define a finite open cover $\bigcup_i \mathcal J_i$ of $\bigcup_{i=1}^{n}S_i$. The fact that a finite union of finite sets is finite implies that $\bigcup_i J_i$ is a finite open cover of $\bigcup_{i=1}^{n}S_i$ $\therefore \bigcup_{i=1}^{n}S_i$ is compact by the definition of a compact topological space.

 

[fresh_42]

Homework Statement:: prove that a union of compact spaces is compact
Relevant Equations:: o.o

Prove that if $X$ is a topological space, and $S_i \subset X$ is a finite collection of compact subspaces, then their union $S_1 \cup \cdots \cup S_n$ is also compact.

$S_i \subset X$ is compact $\therefore \forall S_i, \exists$ a finite open cover $\mathcal J_i=\{U_j\}_{j\in J\subset K} \subset \{U_k\}_{k\in K}=\mathcal K_i$, where $\mathcal K_i$ is an open cover of $S_i$. I define an open cover $\bigcup_i \mathcal K_i$ of $\bigcup_{i=1}^{n}S_i$. Then, an open cover of $\bigcup_{i=1}^{n}S_i$ is simply $\bigcup_i J_i$. The fact that a finite union of finite sets is finite implies that $\bigcup_i J_i$ is a finite open cover of $\bigcup_{i=1}^{n}S_i$ $\therefore \bigcup_{i=1}^{n}S_i$ is compact by the definition of a compact topological space.

The idea is correct. However, you have to start with an arbitrary open cover of all $S_i$. You kick in with finite subcovers of the single $S_i$, which is too late. Where did you get $\mathcal{K}$ from?

 

[PeroK]

Homework Statement:: prove that a union of compact spaces is compact
Relevant Equations:: o.o

Prove that if $X$ is a topological space, and $S_i \subset X$ is a finite collection of compact subspaces, then their union $S_1 \cup \cdots \cup S_n$ is also compact.

Note that it is sufficient to show this for $n = 2$.

 

[docnet]

The idea is correct. However, you have to start with an arbitrary open cover of all $S_i$. You kick in with finite subcovers of the single $S_i$, which is too late. Where did you get $\mathcal{K}$ from?

Thank you! $\mathcal{K}_i$ is an arbitrary open cover of $S_i$. Each of $\mathcal{K}_i$ contains the finite subcover $\mathcal J_i$. I confused the term "finite subcover" with "finite open cover". Is that what you mean to clarify?

Here is my re-attempt:

$\forall S_i\subset X, \exists$ an arbitrary open cover $\mathcal K=\{U_k\}_{k\in K}$. Because $S_i$ is finite, $\exists$ a finite subcover $\mathcal J_i=\{U_j\}_{j\in J} \quad \forall S_i$ s. t. $J\subset K$. Then, I define an arbitray open cover of $\bigcup_{i=1}^{n}S_i$ by writing $\bigcup_i \mathcal K_i$. $\bigcup_{i=1}^{n}S_i$ is finite because a finite union of finite sets is finite. $\therefore \forall K_i, \exists$ a finite subcover $\bigcup_i \mathcal J_i$ Since each of $J_i$ covers all of $S_i$. $\therefore \bigcup_{i=1}^{n}S_i$ is compact by the definition of a compact topological space.

Any suggestions or "neat tricks" would be soo much appreciated :) Thank you!

 

[fresh_42]

No. You have to start at the beginning. Given any open cover $\mathcal{K}$ of $\cup S_i$. Break.

Now you use this to get an open cover of the single $S_i$. Break.

Then you use the compactness condition of the $S_i$, for each $i$. Break.

Finally, you gather all $n$ subcovers to a subcover of $\cup S_i$. Break.

Count the sets in the subcover. (No induction or restriction on $n$ necessary.)

 

[PeroK]

Thank you! $\mathcal{K}_i$ is an arbitrary open cover of $S_i$. Each of $\mathcal{K}_i$ contains the finite subcover $\mathcal J_i$. I confused the term "finite subcover" with "finite open cover". Is that what you mean to clarify?

Here is my re-attempt:

$\forall S_i\subset X, \exists$ an arbitrary open cover $\mathcal K=\{U_k\}_{k\in K}$. Because $S_i$ is finite, $\exists$ a finite subcover $\mathcal J_i=\{U_j\}_{j\in J} \quad \forall S_i$ s. t. $J\subset K$. Then, I define an arbitray open cover of $\bigcup_{i=1}^{n}S_i$ by writing $\bigcup_i \mathcal K_i$. $\bigcup_{i=1}^{n}S_i$ is finite because a finite union of finite sets is finite. $\therefore \forall K_i, \exists$ a finite subcover $\bigcup_i \mathcal J_i$ Since each of $J_i$ covers all of $S_i$. $\therefore \bigcup_{i=1}^{n}S_i$ is compact by the definition of a compact topological space.

Any suggestions or "neat tricks" would be soo much appreciated :) Thank you!

To be honest, you seem to be just throwing symbols around here. I would first try to write a coherent outline of a proof of why the proposition is true.

 

[fresh_42]

To be honest, you seem to be just throwing symbols around here.

This is my impression, too. The crucial point is

Given any open cover $\mathcal{K}$ of $\cup S_i$.

We do not bother the existence of such a cover. We simply assume one. It exists per condition, not because it actually does. If we have an open cover, then we have to find a finite subcover.

You have one cover of the union (per assumption) from which you have to find a cover of all of the $S_i.$ Then your condition kicks in. From that, you have to gather a finite subcover of the union.

 

[docnet]

Thanks @PeroK and @fresh_42 for the advice and patience! It is my first semester of writing these proofs and my brain is ouchy.

Is this sufficient for proof of compactness ??

Suppose there is an open cover $\mathcal V$ of $\bigcup S_i$. Then for every $S_i$ there is an open cover $\mathcal V_i= \mathcal V \cap S_i$. That $S_i$ is compact implies the existence of a subcover $\mathcal U_i$ in every $Vi$ with $n<\infty$ elements. Take the union of all the $U_i$ to construct a subcover of $\bigcup S_i$ which has $i\times n$, a finite number of sets. It follows that that $\bigcup S_i$ is compact

 

[fresh_42]

Thanks @PeroK and @fresh_42 for the advice and patience! It is my first semester of writing these proofs and my brain is ouchy.

Is this sufficient for proof of compactness ??

Suppose there is an open cover $\mathcal V$ of $\bigcup S_i$. Then for every $S_i$ there is an open cover $\mathcal V_i= \mathcal V \cap S_i$. That $S_i$ is compact implies the existence of a subcover $\mathcal U_i$ in every $Vi$ with $n<\infty$ elements. Take the union of all the $U_i$ to construct a subcover of $\bigcup S_i$ which has $i\times n$, a finite number of sets. It follows that that $\bigcup S_i$ is compact

Yes. And you don't even have to go down to $\mathcal{V}\cup S_i$. $\mathcal{V}$ itself is already an open cover of every single $S_i$, because we are in the same topological space $X$. Your proof works for the product topology, here we already have an open cover of the $S_i$ as soon as we have one for $\cup S_i$. The finite subcovers of the single $S_i$, however, are possibly all different. Say we have $m_i$ open sets which cover $S_i$. Then our finite subcover of $\cup S_i$ has $\sum_{i=1}^n m_i$ many open sets, not $i \cdot n$.

It is a good idea for any proof to start with a list of what is given, and what it actually means, i.e. list the definitions. I tried to explain the structures here:
https://www.physicsforums.com/insights/how-most-proofs-are-structured-and-how-to-write-them/

It's not very good, I have to admit, but maybe it helps a bit anyway.

 

[docnet]

Yes. And you don't even have to go down to $\mathcal{V}\cup S_i$. $\mathcal{V}$ itself is already an open cover of every single $S_i$, because we are in the same topological space $X$. Your proof works for the product topology, here we already have an open cover of the $S_i$ as soon as we have one for $\cup S_i$. The finite subcovers of the single $S_i$, however, are possibly all different. Say we have $m_i$ open sets which cover $S_i$. Then our finite subcover of $\cup S_i$ has $\sum_{i=1}^n m_i$ many open sets, not $i \cdot n$.

It is a good idea for any proof to start with a list of what is given, and what it actually means, i.e. list the definitions. I tried to explain the structures here:
https://www.physicsforums.com/insights/how-most-proofs-are-structured-and-how-to-write-them/

It's not very good, I have to admit, but maybe it helps a bit anyway.

Thanks @fresh_42, you are amazin!

So $\mathcal{V}$ is already a cover for every $S_i$ since it is a cover for the union of the $S_i$.

Thank you for the link that explains how to write proofs! I am not sure if I am able to understand it all at this moment, but I will bookmark it and refer to it from time to time!.

 

[docnet]

For completeness, here is the final proof to the problem: prove that the union of compact subspaces is compact.

Suppose there is a cover $\mathcal V$ of $\bigcup S_i$. Then $\mathcal V$ is an open cover for every $S_i$. That $S_i$ is compact implies the existence of a subcover $\mathcal {U}_i$ with $m_i<\infty$ elements. Take the union of $U_i$ to construct a subcover of $\bigcup S_i$, which has $\sum_{i=1}^nm_i$, a finite number of open sets. It follows that $\bigcup S_i$ is compact.

Thanks again for helping me write this (what feels like an elegant) proof! @fresh_42 and @PeroK!

 

[PeroK]

For completeness, here is the final proof to the problem: prove that the union of compact subspaces is compact.

Suppose there is a cover $\mathcal V$ of $\bigcup S_i$. Then $\mathcal V$ is an open cover for every $S_i$. That $S_i$ is compact implies the existence of a subcover $\mathcal {U}_i$ with $m_i<\infty$ elements. Take the union of $U_i$ to construct a subcover of $\bigcup S_i$, which has $\sum_{i=1}^nm_i$, a finite number of open sets. It follows that $\bigcup S_i$ is compact.

Thanks again for helping me write this (what feels like an elegant) proof! @fresh_42 and @PeroK!

I would have preferred a bit more detail to show that you really understand all the steps in this proof. Until you get the hang of proof-writing you should be breaking things down more and justifiying to yourself each step.

Also, I'll do the proof for $n = 2$. Although it's almost as easy to do a direct proof for arbitrary $n$, recognising that $n = 2$ and a bit of induction is sufficient is an important and useful idea.

Let $\mathcal V$ be an open cover for $S_1 \bigcup S_2$. First, we show that $\mathcal V$ is an open cover for both $S_1$ and $S_2$:

Let $x \in S_1$. As $x \in S_1 \bigcup S_2$, there is at least one set in $\mathcal V$ that contains $x$. Therefore, $\mathcal V$ is an open cover for $S_1$. The same argument also applies to $S_2$.

As $S_1$ is compact, $\mathcal V$ has a finite subcover of $S_1$. We'll denote this by $\mathcal V_1$. Likewise we can find $\mathcal V_2$, which is a finite sub-cover for $S_2$.

Now, we define $\mathcal V'$ as all the sets that are either in $\mathcal V_1$ or $\mathcal V_2$ (or both). Finally, we note that $\mathcal V'$ is a finite sub-cover of $\mathcal V$ for $S_1 \bigcup S_2$, which follows directly from the construction of $\mathcal V'$:

1) $\mathcal V'$ is a finite sub-family of $\mathcal V$.

2) $\mathcal V'$ covers $S_1 \bigcup S_2$.

In conclusion, we chose an arbitrary open cover for $S_1 \bigcup S_2$ and constructed a finite sub-cover. It follows that $S_1 \bigcup S_2$ is compact.

A simple inductive argument shows that if the proposition holds for $n = 2$, the proposition holds for any $n$.

I suggest that at this stage you should be attempting this sort of careful step-by-step proof, where you can justify every step.

 

[WWGD]

I suspect you have to cap the cardinality of your spaces. In $$\mathbb R^n$$, a space is compact iff it is closed and bounded. A countably-infinite collection of disjoint compact spaces is not necessarily bounded. I am thinking, e.g., $$ \cup [n, n+1]; n \in \mathbb N$$