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Definition of compact that I'm using: $E$ is compact if each of its open covers has a finite subcover.

Example: Let $X = \mathbb{R}$ with the topology $T = \left \{ U \subset X: U = X \text{ or } 1 \notin U \right \}$. We will show $X$ is compact. Let $C$ be any open cover of $X$. Then there exists $U \in C$ such that $1 \in U$. By definition of $T$ we must have $U = X$. So, $\left \{ U\right \}$ is a finite subcover of $C$. Hence, $X$ is compact.

I'm having trouble understanding the bold there exists. To show compactness, I thought we had to start with an arbitrary open cover and show that EVERY such open cover has a finite subcover? I feel that in this proof we are only finding one such open cover with the property we want. (How would we know that each arbitrary open cover has the property that $1 \in U$?)

Dominated Convergence Theorem
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2 Answers2

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The cover $C$ covers every point of $\mathbb{R}$. In particular the point $1$ must be covered by an open set in the cover.

Sri-Amirthan Theivendran
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The only thing that is said from $C$ is that it is an open cover of $X$.

In spite of this lack of specifications still it is managed to prove that $C$ has a finite subcover.

So for having a finite subcover apparantly it is enough allready to be an open cover of $X$.

In other words: every open cover of $X$ has a finite subcover.

Of even shorter: $X$ is compact.

drhab
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