Suppose $C_1,C_2,...,C_n$ are compact sets on $\mathbb{R}$. Prove that $\underset{i = 1}{\bigcup}^n C_i$ is compact

Question

This was one of the problems on my last exam. The proof I provided was mostly correct, but my professor took off a couple points for a mistake I made. I'll number the lines of the proof for reference.

Proof

1 Let $\mathcal{U}$ be an open cover of $C_n$

2 Then, $\mathcal{U}\cup\{\mathbb{R} \backslash C_n\}$ is an open cover of $C_1,C_2,...,C_{n-1}$

3 Since $C_n$ is compact, there is a finite collection of open intervals $U_1,U_2,...,U_k \in \mathcal{U}$ such that $\underset{i = 1}{\bigcup}^k U_i \supseteq C_n$

4 So $\underset{i = 1}{\bigcup}^k U_i \cup\{\mathbb{R} \backslash C_n\} \supseteq C_i$ for $1 \leq i \leq n$

5 Then $\underset{i = 1}{\bigcup}^k U_i \cup \{\mathbb{R} \backslash C_n\} \supseteq \underset{i = 1}{\bigcup}^n C_i$

6 Thus, $\underset{i = 1}{\bigcup}^n C_i$ is compact

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So my professor thought the proof was mostly correct except for Line 2. On Line 2, when I said that

"$\mathcal{U}\cup\{\mathbb{R} \backslash C_n\}$ is an open cover of $C_1,C_2,...,C_{n-1}$," he said, "Not necessarily."

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So why isn't $\mathcal{U}\cup\{\mathbb{R} \backslash C_n\}$ an open cover of $C_1,C_2,...,C_{n-1}$ ?

We know that $\{\mathbb{R} \backslash C_n \}$ covers all of $C_1, C_2, ..., C_{n-1}$ if all of the $C_i$ are disjoint.

If the $C_i$'s are not disjoint (meaning at least one $C_i$ has $C_i \cap C_n \neq \emptyset$), then $\mathcal{U} \cup \{\mathbb{R} \backslash C_n\}$ covers all of them. What am I missing here?

Answer 1

There is nothing wrong with Line $2$. $C_n\subseteq\bigcup\mathscr{U}$, so $(\Bbb R\setminus C_n)\cup\bigcup\mathscr{U}=\Bbb R$, and therefore $\mathscr{U}\cup\{\Bbb R\setminus C_n\}$ is in fact an open cover of $\Bbb R$. However, I would have given very little credit for your argument: you have not shown that an arbitrary open cover of $\bigcup_{k=1}^nC_k$ has a finite subcover.

Start with an arbitrary open cover $\mathscr{U}$ of $\bigcup_{k=1}^nC_k$. $\mathscr{U}$ is an open cover of $C_k$ for each $k\in\{1,\ldots,n\}$, so for each $k\in\{1,\ldots,n\}$ there is a finite subset $\mathscr{U}_k$ of $\mathscr{U}$ that covers $C_k$. Let $\mathscr{V}=\bigcup_{k=1}^n\mathscr{U}_k$; clearly $\mathscr{V}$ is a finite subset of $\mathscr{U}$ that covers $\bigcup_{k=1}^nC_k$, which is therefore compact.

Answer 2

$\mathcal U\cup\{\Bbb R\setminus C_n\}$ covers all the $C_i$-s because it covers $\Bbb R$.

Your mistake lies else where. Other than the small imprecision of you saying that $\mathcal U$ is an open cover, but then claiming to take "intervals" from $\mathcal U$ (in fact, $\mathcal U$ may not contain intervals at all), there's the catastrophic issue that you have only proved that all covers coming from a construction such as yours admit a finite subcover: effectively, just the covers $\mathcal V$ of $\bigcup_{i=1}^n C_i$ such that $\Bbb R\setminus C_n\in \mathcal V$ (or, in a broader sense, the ones such that there is some $i$ such that $\Bbb R\setminus C_i\in\mathcal V$). But you have to prove existence of a finite subcover for all open covers of $\bigcup_{i=1}^n C_i$.

Answer 3

Further to my comment, this is what a proof would look like:

Let $\{U_{j}\}_{j}$ be an open cover of $\bigcup_{i=1}^{n} C_{n}$. Then $\{U_{j}\}_{j}$ covers each $C_{i}$. Since each $C_{i}$ is compact, you can find $U^{i}_{1}, \dots, U^{i}_{k_{i}}$ that cover $C_{i}$. Now consider the cover $$ \bigcup_{i=1}^{n} (U^{i}_{1} \cup \dots \cup U^{i}_{k_{i}}) $$ This has only finitely many sets from your original cover and covers all $C_{i}$, so it covers $\bigcup_{i=1}^{n} C_{i}$, and we're done.

The conceptual error you made in your answer is that you started with an open cover of $C_{n}$ (just one of the sets). When using this definition (open cover, finite subcover) to show that something is compact, the proof will (pretty much) always go this way:

1. Pick an arbitrary open cover of the set that you want to prove is compact (!!!)
2. Use the assumptions given in the question to try to find a subset that covers the set