= Compact space
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= Quasi-compact topological space
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= Quasi-compactness
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= Quasi-compact
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= Compact subset
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= Compact
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= Compact set
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= Compactness
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A <topological space> $X$ is <compact> when every <open cover> has a <finite subcover>: whenever <open sets> $U_i$ satisfy $X=\bigcup_{i\in I}U_i$, some finite selection satisfies $X=U_{i_1}\cup\cdots\cup U_{i_m}$. A <subset> $K$ of an ambient <topological space> is <compact> when its <subspace topology> is <compact>; equivalently, every <open cover> of $K$ by ambient <open sets> has a <finite subcover>. The word \i[every] matters: finding one <finite subcover> of one <open cover> is not enough to establish <compactness>. https://ocw.mit.edu/courses/18-101-analysis-ii-fall-2005/babd982be745679b6d691f78b1c18f53_lectures.pdf[MIT Analysis II, section 1.6 and Definition 1.23]{ref}
The <closed interval> $[0,1]$ is <compact> by the <Heine-Borel theorem>. For a concrete <open cover>, take the three <open intervals> $(-0.1,0.45)$, $(0.35,0.8)$, and $(0.7,1.1)$. Their <union> is $(-0.1,1.1)$, so they cover every point of $[0,1]$, including both endpoints. The overlaps have lengths $0.1$ and $0.1$, leaving no gaps. Each member is necessary for this particular <open cover>: removing them respectively leaves $0$, $0.5$, or $1$ uncovered.
\Image[/compact-open-cover.png]
{title=A <finite subcover> of a <closed interval>, and an <open cover> with no <finite subcover>}
Above, hollow markers denote the excluded endpoints of the covering <open intervals>, while the solid endpoints belong to the covered <closed interval>. Below, each displayed <open interval> is one member of an infinite <open cover> of $(0,1)$; finitely many rows always leave points near zero uncovered. The rows are vertically separated for readability and all refer to the same <real line>.
The <open interval> $(0,1)$ is a <bounded set> but is not <compact>: the <open sets>
$$
U_n=(1/n,1),\qquad n\geq2,
$$
give an <open cover> of $(0,1)$, since every $0<x<1$ has $1/n<x$ for sufficiently large $n$. These are also <open sets> for the <subspace topology> on $(0,1)$. Any nonempty finite selection has a largest index $N$, so its <union> is just $(1/N,1)$ and misses the point $1/(N+1)$, which still belongs to $(0,1)$. Thus no <finite subcover> exists, however many members we select. The same family is not an <open cover> of $[0,1]$, since it misses both endpoints.
In a <metric space>, <compactness> is equivalent to <sequential compactness>: every <sequence> in the <subset> has a <convergent subsequence> whose <limit of a sequence> belongs to that <subset>. This equivalence does not extend to all <topological spaces>. For example, $x_n=1/n$ for $n\geq2$ has <limit of a sequence> zero in $[0,1]$. The same <sequence> lies in $(0,1)$, but every <subsequence> still has <limit of a sequence> zero on the <real line>, so no <subsequence> converges to a point of $(0,1)$. Merely approaching a <boundary> is harmless when the required point remains available. https://ocw.mit.edu/courses/18-100b-analysis-i-fall-2010/b7780b2a0629359b11c5de998c72a147_MIT18_100BF10_compactness.pdf[MIT Analysis I, Compactness vs. Sequential Compactness, Theorem 1 and its metric-space proof]{ref}
\Image[/compactness-missing-endpoint.png]
{title=The same <sequence> approaches an included endpoint in a <closed interval> and a missing endpoint in an <open interval>}
Both panels show the <sequence> $x_n=1/n$ for $n\geq2$ approaching zero. In the <closed interval>, zero is an allowed <limit of a sequence>; in the <open interval>, the hollow marker identifies the missing point. This single <sequence> disproves <sequential compactness> of $(0,1)$; <compactness> of $[0,1]$ still requires control of every <open cover> or every <sequence>, supplied by the <Heine-Borel theorem>.
An <infinite set> with the <discrete topology> similarly fails <compactness>, using the <open cover> by all its <singletons>. Every <finite set> is <compact> in any <topology>: select one covering <open set> for each point. The <empty set> is <compact> because the empty selection is a <finite subcover>. This <open cover> definition does not require the <topological space> to be <Hausdorff>; in <algebraic geometry> it is often called <quasi-compactness>.
The beauty of <compactness> is that information valid around each point can often be controlled using finitely many regions. A <finite subcover> lets us combine finitely many local bounds, or choose the smallest of finitely many positive tolerances, to obtain control across the whole <compact space>. This is why <compactness> repeatedly turns approximation into existence and why it is so useful in <optimization> through the <extreme value theorem>.
Every <closed set> in a <compact space> is <compact>, and a <continuous image of a compact space> is <compact>. The latter proof pulls an <open cover> back by <inverse images>, selects a <finite subcover>, and maps forward. This makes <compactness> useful for proving existence: the <extreme value theorem> guarantees that a <continuous function> from a nonempty <compact space> to the <real line> attains both a <minimum> and a <maximum>. For example, $q(x)=x(1-x)=1/4-(x-1/2)^2$ on $[0,1]$ has <maximum> $1/4$ at $1/2$ and <minimum> zero at its endpoints. On $(0,1)$ its <infimum> is still zero but is never attained. https://ocw.mit.edu/courses/18-101-analysis-ii-fall-2005/babd982be745679b6d691f78b1c18f53_lectures.pdf[MIT Analysis II, Theorems 1.24-1.25]{ref}
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