What is wrong with this argument that closed interval [0, 1] is not compact?

Question

I am a student majoring engineering.

I am studying real analysis with textbook 'Measure and Integral' by Wheeden and Zygmund.

This book defined compact like the following:

$E$ is compact if every open cover of $E$ has a finite subcover.

By the definition $[0, 1]$ is not compact.

However, by the Heine-Borel theorem, $[0, 1]$ is compact.

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Let me prove why $[0, 1]$ is not compact by the definition.

According to the definition, it is enough to show an open cover of $E$ having infinite subcover.

If $C=\{U_\alpha:\alpha\in \mathbb{N}\}$ is an indexed family of sets $\displaystyle U_\alpha=\left(-1+\frac{1}{n}, 2\right)$, then $C$ is a cover of $[0, 1]$ because $\displaystyle [0,1] \subseteq \bigcup\limits_{\alpha \in \mathbb{N}} {\mathop U\nolimits_\alpha }$.

This $C$ has infinite subcovers like $C=\{U_{2\alpha}:\alpha\in \mathbb{N}\}$

Can someone teach me what is my fault?

Answer 1

By your definition of compactness, $[0,1]$ is compact. You misunderstood and the proof is incorrect.

You should try to prove that if you take an arbitrary infinite cover of E, you can find within that cover a finite sub-cover. What you did was: you picked one particular infinite cover of E and then you showed that it has an infinite sub-cover. That doesn't do the work here.

Answer 2

Your "proof" is wrong.

The definition says that for every cover there exists a finite subcover. In your post there is an example of an infinite subcover of $\{U_\alpha\}$.

By following your reasoning every subset $U$ of a topological space would be non-compact: pick a "redundant" infinite open cover of $U$ (you can always find it) and remove one set. You have an infinite subcover.

Answer 3

As said in the other answers, you misunderstood a point of the definition.

It is actually tricky to prove that $[0,1]$ is compact. You have to consider an arbitrary cover of $[0,1]$, let's call it $\mathcal{U}$. Then, consider the set $A=\{x\in [0,1]$ such that $\mathcal{U}$ has a finite subcover covering $[0,x]\}$. The proof itself is done in three steps: First, prove that $A$ is non empty. Then prove that sup$A\in A$. And finally, prove that sup$A=1$. Maybe it doesn't look so hard, but the third step is not trivial.

Answer 4

To show a set is not compact, you have to exhibit an open cover where every subcover is infinite (not just one). Alternatively, you have to exhibit an open cover where every finite subset fails to be a cover (so every subcover, if it exists, must by infinite).

So, for example, $(0,1)$ is not compact, because it has the infinite cover $\{ (1/n, 1) : n \in \mathbb{N}\}$; but if I pick any finite subset of that cover $$\{(1/n_{0}, 1), (1/n_{1}, 1), \ldots, (1/n_{m}, 1)\},$$ and I let $$m = \sup\{n_{0}, n_{1}, \ldots, n_{m}\},$$ then $$\bigcup \{(1/n_{0}, 1), (1/n_{1}, 1), \ldots, (1/n_{m}, 1)\} = (1/m, 1)$$ which does not include the whole set $(0,1)$, so my finite set is not a subcover.

To prove $[0,1]$ is not compact, you would have to show similarly that every finite subset of your cover is not a subcover; but since your cover does have (some) finite subcovers (as well as infinite ones), you can't do that.