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Let $X,Y$ be topological spaces, let $X$ be compact, and let $f:X\to Y$ be a continuous map. Then its image $f(X)$ is compact with the subspace topology inherited from $Y$. More generally, a continuous map sends every compact subset $K\subseteq X$ to a compact subset $f(K)\subseteq Y$, by applying the same statement to its restriction to $K$. Neither $X$ nor $Y$ needs to be Hausdorff.^[ref]

For example, the polynomial function $f(x)=x^2$ sends the compact closed interval $[0,1]$ onto $[0,1]$. The open intervals $V_1=(-1/10,16/25)$ and $V_2=(9/25,11/10)$ give an open cover of the image. Their inverse images in the domain areBoth are open sets in the subspace topology of $[0,1]$, even though neither is an open set in the ambient real line. They cover the domain, so the corresponding two target open sets cover the image.

Matching colors identify the target open sets and their inverse images. Filled endpoints are included and hollow endpoints excluded. This particular open cover illustrates the mechanism; compactness requires a finite subcover for every open cover.

For the general proof, write an arbitrary open cover of $f(X)$ as $\{f(X)\cap V_i:i\in I\}$ with each $V_i$ an open set in $Y$. By continuity, the inverse images $f^{-1}(V_i)$ form an open cover of $X$. Its compactness provides indices $i_1,\dots ,i_m$ such thatThis gives the required finite subcover of $f(X)$. The argument uses inverse images because continuous maps need not send open sets to open sets. For the empty set, the empty selection suffices.

The image of a compact Hausdorff space under a continuous map need not itself be Hausdorff. Take $D=\{0,1\}$ with the discrete topology and $S=\{0,1\}$ with the Sierpiński space topology $\{\varnothing ,\{1\},S\}$. The identity map $D\to S$ is a continuous bijection, since the inverse image of every target open set is an open set in $D$. Its image is compact, but the two points of $S$ have no disjoint open neighborhoods. If the target $Y$ is Hausdorff, its subspace topology does make $f(X)$ Hausdorff, and the compact subset of a Hausdorff space theorem also makes $f(X)$ closed in $Y$.

The practical payoff is existence: the extreme value theorem uses the compact image of a nonempty compact space under a real-valued continuous function to obtain an attained minimum and maximum. When $f$ is also a bijection onto a Hausdorff target, the compact-to-Hausdorff continuous bijection theorem supplies continuity of its inverse without a separate formula.

Every subset $A$ of a cofinite topological space is a compact subset. To see the finite reduction directly, take an open cover of a nonempty $A$ by open sets of $X$ and choose one member $U$ containing a point of $A$. Since $U$ has finite complement, the set $A\setminus U$ is finite. Choose a covering member for each of those remaining points. Those members together with $U$ are a finite subcover. The empty set is already compact.

Equivalently, the subspace topology on $A$ is itself cofinite: for finite $F\subseteq X$, the trace of $X\setminus F$ is $A\setminus (F\cap A)$. The argument applies again inside $A$. A concrete open cover $\{X\setminus \{k\}:k\in X\}$ of $X={\mathbb{N}}_{>0}$ needs only the two members $X\setminus \{1\}$ and $X\setminus \{2\}$ when these labels are distinct: each fills the other's single gap.

On $X={\mathbb{N}}_{>0}$, the subset of even integers is compact but not closed, because it is infinite and proper. A compact subset of a Hausdorff space is guaranteed to be closed, whereas compactness without the Hausdorff space assumption does not suffice. Changing the topology changes compactness: the same infinite set with its discrete topology has the singleton open cover, which has no finite subcover. The finite-exception argument is special to finite complements; in the cocountable topology, compact subsets of a cocountable space must instead be finite.

A Hausdorff space is a topological space in which any two distinct points have disjoint open neighborhoods. Let $K$ be a compact subset of a Hausdorff space and $x\notin K$. For each $y\in K$, choose disjoint open neighborhoods $U_y$ of $y$ and $W_y$ of $x$. The $U_y$ cover $K$, so compactness gives a finite subcover $U_{y_1},\dots ,U_{y_m}$. Their paired intersection $W={\bigcap }_{r=1}^m{W}_{y_r}$ is an open neighborhood of $x$ disjoint from $K$. Thus every point of its complement has an open neighborhood in the complement, proving every compact subset of a Hausdorff space is closed. The empty set case is immediate.

In the cocountable topology on an uncountable set, any two nonempty open sets intersect: their combined complements are countable and cannot exhaust $X$. Thus the space is not Hausdorff.

Nevertheless every compact subset is closed here. In fact only finite subsets are compact. To prove this, suppose $K$ is infinite and choose distinct points $d_1,d_2,\dots \in K$. The open setsform an increasing open cover of $X$, and hence of $K$. Any finite selection is contained in $U_N$ for its largest index $N$, which misses $d_N\in K$. So $K$ is not compact. Finite sets are compact in any topological space, and in this cocountable topology they are closed because their complements are open. This proves the claimed behavior of compact subsets of a cocountable space.

For every $k\in K$, the Hausdorff condition gives disjoint open neighbourhoods $U_k$ of $k$ and $V_k$ of $x$. The $U_k$ form an open cover of $K$, so choose a finite subcover $U_{k_1},\dots ,U_{k_m}$. PutThen $U,V$ are open sets, $K\subseteq U$ and $x\in V$. If a point belonged to $U\cap V$, it would belong to some $U_{k_j}$ and to $V_{k_j}$, contrary to their choice. ThusThe finiteness matters: a finite intersection of open sets is open. This is separation of a point and a compact set. If $K=\varnothing$, use $U=\varnothing$, $V=X$.

Every $x\in X\setminus K$ consequently has an open neighbourhood $V$ contained in $X\setminus K$. The complement is open, so a compact subset of a Hausdorff space is closed.

The conclusion can fail without the Hausdorff condition. Take the two-point set $X=\{0,1\}$ with the indiscrete topology $\{\varnothing ,X\}$. The singleton $K=\{0\}$ is compact, since one member of any open cover suffices. It is not closed, because its complement $\{1\}$ is not open. The two points cannot have disjoint open neighbourhoods, so this example is indeed not Hausdorff.

Write $K={\bigcap }_{i\in I}{K}_i$. Each $K_i$ is closed in $X$, by the compact subset of a Hausdorff space theorem. Therefore $K$ is closed, and is compact because $X$ is compact. The nonempty finite intersections give the finite intersection property, so part (a)(ii) also yields $K\ne \varnothing$.

Suppose $K$ were not connected. Part (c) would give disjoint open sets $U,V$ containing $K$ in their union and each meeting $K$. LetThis is a closed subset of the compact space $X$, and $C\cap {\bigcap }_{i\in I}{K}_i=\varnothing$. Apply the contrapositive of the finite intersection characterization of compactness to the family of closed sets $\{C\}\cup \{K_i:i\in I\}$. Some finite subfamily has empty intersection. Since every finite intersection of just the $K_i$ is nonempty, that subfamily must include $C$. Thus for some finite $J\subseteq I$,But $K\subseteq K_J$ and $K$ meets both $U$ and $V$. Therefore $K_J\cap U$ and $K_J\cap V$ are nonempty disjoint relatively open sets covering $K_J$, contradicting its assumed connectedness. This provesThis is connectedness of an intersection from connected finite intersections. If $I=\varnothing$, its intersection is $X$; the stated hypothesis for the finite set $J=\varnothing$ already makes $X$ nonempty and connected, so that case is included.

Suppose $K\setminus \{a\}$ were compact. By the compact subset of a Hausdorff space theorem, it would be closed in $X$, hence closed in $K$. The singleton $\{a\}$ is also closed in the Hausdorff space $X$: each other point has an open neighbourhood avoiding $a$.

Consequently both $\{a\}$ and $K\setminus \{a\}$ are clopen in $K$. They are disjoint, nonempty, and cover $K$, giving a separation of a topological space in contradiction to the connectedness of $K$. Thus $K\setminus \{a\}$ is not compact.