I am trying to show the following:
Let $X,Y$ be compact topological spaces. Then the disjoint union $X \sqcup Y$ is compact.
In my course, we defined the disjoint union of an arbitrary family $\{X_i\}_{i \in I}$ of spaces as
$$
\bigsqcup_{i \in I}X_i := \{ (i,x) \mid i \in I,\; x \in X_i \},
$$
where we additionally have canonical inclusion maps
$$
\iota_j : X_j \to \bigsqcup_{i \in I}X_i, \quad x \mapsto (j,x).
$$
So in my case, we consider maps $\iota_1 : X \to X \sqcup Y$ and $\iota_2 : Y \to X \sqcup Y$.
My attempt: Let $\bigcup_{i\in I}U_i$ be an open covering of $X \sqcup Y$, i.e. $U_i \subseteq X \sqcup Y$ is open for all $i \in I$, meaning $\iota_1^{-1}(U_i)\subseteq X$ and $\iota_2^{-1}(U_i)\subseteq Y$ are open. Then,
$$
X = \iota_1^{-1}\left( \bigcup_{i\in I}U_i \right) = \bigcup_{i \in I}\iota_1^{-1}(U_i),
$$
and since $X$ is compact, there exists $J_1 \subseteq I$ with $\#J_1<\infty$, such that $X = \bigcup_{i \in J_1}\iota_1^{-1}(U_i)$. Similarly, there exists finite $J_2 \subseteq I$, such that $Y = \bigcup_{i \in J_2}\iota_2^{-1}(U_i)$. Then, we get that
$$
X \sqcup Y = \bigcup_{i \in J_1 \cup J_2}U_i,
$$
and $\#(J_1 \cup J_2)<\infty$, so $X \sqcup Y$ is compact.
My question is whether or not this proof is valid. I feel like I'm at least going in the right direction, but I'm not entirely sure that all steps are fully rigorous.
Best Answer
What you have written is a good and formal proof. Here's another way to think about it: The maps $\iota_1$ and $\iota_2$ are not just injective but they're actually embeddings, i.e., $\iota_1(X)$ is homeomorphic to $X$ and similarly for $Y$.
Under this, you can identify $X$ and $Y$ as subspaces of $X \sqcup Y$. In particular, as subsets. The way of writing the proof is now a bit simplified. Given a cover $\{U_i\}_{i \in I}$ of $X \sqcup Y$, it is in particular, a cover of $X$ and $Y$ both.
By compactness, you can cover each of $X$ and $Y$ individually by finitely many $U_i$. Together, they must cover the union, which is $X \sqcup Y$.
In general, this argument shows that given any topological space $Z$ and compact subspaces $X, Y \subset Z$, the subspace $X \cup Y$ is compact. (Even if $X$ and $Y$ are not disjoint.)