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.


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

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."

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?

  • 3
    $\begingroup$ As far as I can see the 'proof' is completely wrong. You did not even start with an open cover of $\cup C_i$. $\endgroup$ Nov 29, 2020 at 23:13
  • 2
    $\begingroup$ For starters, to show that $\bigcup_{i=1}^{n} C_{i}$ is compact, you should start with an open cover of $\bigcup_{i=1}^{n} C_{i}$ and not $C_{n}$. $\endgroup$
    – gtoques
    Nov 29, 2020 at 23:14
  • $\begingroup$ My argument for the proof is that if there exists a finite open cover $K \supseteq C_i$ for any $1 \leq i \leq n,$ then $K \subseteq \cup C_i$. Is that not true? $\endgroup$ Nov 29, 2020 at 23:15
  • $\begingroup$ @ManCheese if a set $K$ contains all $C_{i}$ then it surely contains $\bigcup_{i=1}^{n} C_{i}$, but how does that help you? Also, merely showing the existence of a finite open cover is not what compactness is ($\mathbb{R}$ is open, so technically all sets have a finite open cover), the point is to show that ANY open cover admits a finite open subcover $\endgroup$
    – gtoques
    Nov 29, 2020 at 23:27

3 Answers 3


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.


$\mathcal U\cup\{\Bbb R\setminus C_n\}$ 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 $\bigcup_{i=1}^n C_i$.

  • $\begingroup$ I suppose I could have left out the term “interval” since it wasn’t of paramount importance as far as where I was concerned. In any case, I should have just said “open set” instead of “open interval” so as to remove any issues. Furthermore, my proof has far more flaws than you’re giving me credit for, as the answers of others has shown $\endgroup$ Nov 29, 2020 at 23:29
  • $\begingroup$ @ManCheese Sorry, I had misread your comment. $\endgroup$
    – user239203
    Nov 29, 2020 at 23:45

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

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