Proving that the union of compact sets is compact (in $\mathbb{R}$)

Question

Proof

Let $\{C_i\}_{i=1}^n$ be a collection of nonempty compact sets in $\mathbb{R}$

Let $\mathcal{U}$ be an open cover of $\displaystyle\bigcup_{i=1}^n C_i$

Since each $C_i$ is compact, there are finitely many open sets $O_1,O_2,…,O_m \in \mathcal{U}$ such that $\displaystyle\bigcup_{j=1}^m O_j \supseteq C_i$

Since the finite union of finite sets is finite, there exists an finite subcover of $\mathcal{U}$, $\mathcal{U}_o$ such that $\mathcal{U}_o \supseteq\displaystyle\bigcup_{i=1}^n C_i$.

Thus, $\displaystyle\bigcup_{i=1}^n C_i$ is compact. $\hspace{1cm} \square$

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Did I leave anything out or did I make any jumps in my logic? Let me know.

Answer 1

You are on the good tract, you just have to consider $\cup O_{i_n}^i$

where $C_i$ is a subset of the union of open subsets $O^i_{i_1},...,O^i_{i_{n_i}}$.