Would the following proof be wrong? (About the intersections of compact subsets)

Question

Let $X$ be a topological space, and let $\{K_\alpha\}_{\alpha\in A}$ be a family of closed compact subsets of $X$. Show that $\bigcap_{\alpha\in A} K_\alpha$ is compact.

Proof:

Let $\mathcal{T}$ be the given topology on $X$. And let $\mathcal{T}_\alpha$ be the corresponding subspace topology on $K_\alpha$.

Let $K=\bigcap_{\alpha\in A} K_\alpha$ and let $\mathcal{T}_K$ be its subspace topology.

First of all we note that $K$ is closed since it is an intersection of closed sets.

Now, pick any family of closed sets $F:=\{F_\beta\}_{\beta\in B}$ of $K$ such that it has the finite intersection property. We want to show that $\bigcap F\neq\emptyset$.

For each $\beta\in B$, since $F_\beta$ is closed in $K$, $K\setminus F_\beta$ is open in $K$, and hence by the definition of the subspace topology on $K$, $F_\beta=K\setminus(U_i\cap K)=(K\setminus U_i)\cup(K\setminus K)=K\setminus U_i$ for some $U_i\in\mathcal{T}$.

Then, for each $\alpha\in A$, \begin{equation*} K_\alpha\setminus F_\beta=K_\alpha\setminus(K\setminus U_i)=K_\alpha\cap(K\cap U_i')'=K_\alpha\cap(K'\cup U_i)=(K_\alpha\cap K')\cup(K_\alpha\cap U_i) \end{equation*} is open in $K_\alpha$, since $K'$ and $U_i$ are open in $X$, and hence $K_\alpha\cap K'$ and $K_\alpha\cap U_i$ are open in $K_\alpha$ by the definition of subspace topology. (note: we denoted $A'$ be the complement of $A$ in $X$ where $A$ is a subset of $X$.)

Therefore the family $F$ is a family of closed sets in $K_\alpha$ such that it has the finite intersection property for each $\alpha\in A$.

\Then Since each $K_\alpha$ is compact, we conclude that $\bigcap F\neq\emptyset$.

Would this proof be false?

Answer 1

The intersection can easily be empty: $$[0,1]\cap [2,3]=\emptyset.$$ The family having non-empty intersection requires an assumption you have not specified.

However, this is unnecessary. If the intersection is empty you are done. If not, pick an open cover of the intersection. Argue that if it had no finite subcover then neither do the $K_\alpha$.

Answer 2

Let it be that $L$ is a compact space and that $K$ is a closed subset of $L$.

To be shown is that $K$ is compact.

Let $(U_i)_{i\in I}$ be a family of open sets in $L$ with $K\subseteq\bigcup_{i\in I}U_i$.

Also $U:=L-K$ is open and $L=U\cup\bigcup_{i\in I}U_i$. Then a finite set $J\subseteq I$ exists with $L=U\cup\bigcup_{i\in J}U_i$ and consequently we have $K\subseteq\bigcup_{i\in J}U_i$.

It has been shown that the original open cover of $K$ has a finite subcover and this can be done for every open cover of $K$. In other words: $K$ is compact.

If in your question $A\neq\emptyset$ then this can be applied on $L=K_{\beta}$ for some $\beta\in A$ and $K=\bigcap_{\alpha\in A}K_{\alpha}$.