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A finite subcover is a subcover with finitely many members, including possibly none when $K$ is the empty set. Requiring a finite subcover for every open cover is the definition of compactness.

In the subspace topology on $S$, a subset $A$ is open in $S$ precisely when $A=S\cap O$ for some open set $O\subseteq \mathbb{C}$. A subset $B$ is closed in $S$ precisely when $S\setminus B$ is open in $S$, equivalently $B=S\cap F$ for some closed set $F\subseteq \mathbb{C}$.

A compact space is one for which every open cover has a finite subcover. Here the cover may use relatively open subsets of $S$, or equivalently ambient open subsets covering it. The Heine-Borel theorem states thatClosedness and boundedness here refer to the ambient complex plane, not merely closedness in $S$ itself.

Let $(U_i{)}_{i\in I}$ be an open cover of $f(X)$, with the sets open in $Y$. The preimages $f^{-1}(U_i)$ form an open cover of $X$, because $f$ is continuous. By compactness, finitely many of these preimages cover $X$. Their corresponding sets $U_i$ cover $f(X)$, providing a finite subcover. Hence the continuous image is compact.

A subset $A$ is compact when every open cover of $A$ has a finite subcover. This may be stated using sets open in the ambient space or using the subspace topology on $A$; the definitions agree because every subspace-open set is the intersection of $A$ with an ambient-open set.

If $m=\min A$, any open cover contains a member containing $m$. Such a member is either $\mathbb{R}$ or a ray $(a,\infty )$ with $a<m$. It therefore contains every element of $A$, so that single member is already a subcover.

Conversely suppose $A$ is nonempty and has no smallest element. Thenis an open cover of $A$: for any $x\in A$, the lack of a minimum gives an $a\in A$ below $x$. A nonempty finite subcollection has union $(a_0,\infty )$, where $a_0$ is the smallest of its thresholds. It misses $a_0\in A$. The empty subcollection cannot cover nonempty $A$ either. Hence there is no finite subcover, and $A$ is not compact. Thus compactness is equivalent to having a smallest element.

An open cover of a topological space is a collection of open subsets whose union is the whole space. A compact space is one for which every open cover admits a finite subcover. The subspace topology on $A\subseteq Y$ consists of sets $A\cap O$ with $O$ open in $Y$.

Let $\{U_j\}$ cover $f(X)$ by relatively open sets, and write $U_j=f(X)\cap O_j$ with $O_j$ open in $Y$. By continuity, $\{f^{-1}(O_j)\}$ is an open cover of $X$. A finite subcollection covers $X$ by compactness, and its corresponding $U_j$ cover $f(X)$. Thus a continuous image of a compact space is compact.

For an image that is not closed, take $X$ to be a singleton and $Y=\{0,1\}$ with the indiscrete topology. Send the sole point to $0$. This map is continuous and $X$ is compact, while $f(X)=\{0\}$ is not closed in $Y$, whose only closed sets are $\varnothing$ and $Y$. A Hausdorff hypothesis on the target would exclude this example.

Every $K_n$ is closed in the Hausdorff ambient space, because it is compact. Inside the compact space $K_1$, these sets have the finite intersection property: a finite intersection equals the member with largest index and is nonempty. If the total intersection were empty, their open complements would cover $K_1$; a finite subcover would contradict that property. Thus

Every $K_n$ is closed in the Hausdorff ambient space, because it is compact. Inside the compact space $K_1$, these sets have the finite intersection property: a finite intersection equals the member with largest index and is nonempty. If the total intersection were empty, their open complements would cover $K_1$; a finite subcover would contradict that property. Thus

An open cover of a topological space is a collection of open subsets whose union is the whole space. A compact space is one for which every open cover admits a finite subcover. The subspace topology on $A\subseteq Y$ consists of sets $A\cap O$ with $O$ open in $Y$.

Let $\{U_j\}$ cover $f(X)$ by relatively open sets, and write $U_j=f(X)\cap O_j$ with $O_j$ open in $Y$. By continuity, $\{f^{-1}(O_j)\}$ is an open cover of $X$. A finite subcollection covers $X$ by compactness, and its corresponding $U_j$ cover $f(X)$. Thus a continuous image of a compact space is compact.

For an image that is not closed, take $X$ to be a singleton and $Y=\{0,1\}$ with the indiscrete topology. Send the sole point to $0$. This map is continuous and $X$ is compact, while $f(X)=\{0\}$ is not closed in $Y$, whose only closed sets are $\varnothing$ and $Y$. A Hausdorff hypothesis on the target would exclude this example.

A subcover of an open cover $\{U_i:i\in I\}$ of a subset $K$ is a selection of its members whose union still contains $K$. The selected open sets retain their original sizes; selecting a subcover does not mean replacing them by smaller open sets.