= Continuous image of a compact space
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>.<Point-Set Topology by Romyar Sharifi/Chapter 3: Connected and compact spaces>[Lemma 3.2.5 and Proposition 3.2.9]{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 are
$$
f^{-1}(V_1)=[0,4/5),\qquad f^{-1}(V_2)=(3/5,1].
$$
Both 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>.
\Image[/continuous-square-compact-image-open-cover.png]
{title=An <open cover> of the <image> of a <continuous map> and its <inverse images> on a <compact> <closed interval>}
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 $i_1,\ldots,i_m$ such that
$$
X=\bigcup_{j=1}^{m}f^{-1}(V_{i_j})\quad\Longrightarrow\quad f(X)\subseteq\bigcup_{j=1}^{m}V_{i_j}.
$$
This 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.
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