Proof of Compact Subset Criterion via Open Covers in the Ambient Space

We prove $(1)\Rightarrow(2)$ and $(2)\Rightarrow(1)$.

$(1)\Rightarrow(2)$ Assume that $A$ is compact in $X$. Let $(U_i)_{i\in I}$ be an open cover of $A$ in $X$. For each $i\in I$, define

Since each $U_i$ is open in $X$, the set $W_i$ is open in the subspace topology on $A$ by Subspace Topology. Also,

by the definition of open cover, so every point of $A$ lies in some $W_i$. Thus $(W_i)_{i\in I}$ is an open cover of the topological space $A$ equipped with its subspace topology.

Because $A$ is compact in $X$, the definition Compact Topological Space and Compact Subset yields a natural number $n\in\mathbb{N}$ and indices $i_1,\dots,i_n\in I$ such that

Since $W_{i_k}\subseteq U_{i_k}$ for each $k$, it follows that

Therefore $(U_{i_1},\dots,U_{i_n})$ is a finite subcover of $A$ in $X$.

$(2)\Rightarrow(1)$ Assume now that condition $(2)$ holds. To prove that $A$ is compact in $X$, by Compact Topological Space and Compact Subset we must show that every family of sets open in the subspace topology on $A$ whose union contains $A$ admits finitely many members whose union still contains $A$.

Let $I$ be a set, and let $(W_i)_{i\in I}$ be a family of subsets of $A$ such that each $W_i$ is open in the subspace topology on $A$ and

By Subspace Topology, for each $i\in I$ there exists an open set $U_i\subseteq X$ such that

Then $(U_i)_{i\in I}$ is an open cover of $A$ in $X$, because each $U_i$ is open in $X$ and every point of $A$ belongs to some $W_i\subseteq U_i$.

By condition $(2)$, there exist a natural number $n\in\mathbb{N}$ and indices $i_1,\dots,i_n\in I$ such that

Intersecting with $A$ gives

Hence the original family of open sets in the subspace topology has finitely many members covering $A$. By Compact Topological Space and Compact Subset, the subset $A$ is compact in $X$.