The theorem connecting the concept of compactness with boundedness and closedness.

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**Theorem 1** (Theorem of Heine-Borel). *For a subset
\(C\subset\mathbb{K}\), the following
statements are equivalent:*

*\(C\) is
compact;*

*\(C\) is bounded and
closed.*

*Proof:*

“(i)\(\Rightarrow\)(ii)”: Let \(C\) be compact.

Let \((a_n)_{n\in\mathbb{N}}\) be a
convergent sequence in \(\mathbb{K}\)
with \(a_n\in C\) and \(a:=\lim_{n\rightarrow\infty} a_n
\in\mathbb{K}\). Since \(C\) is
compact, there is a subsequence \((a_{n_k})_{k\in\mathbb{N}}\) such that
\(b:=\lim_{k\rightarrow\infty} a_{n_k}\in
C\). By the theorem on the convergence of subsequences, every
subsequence of a convergent sequence is also convergent having the same
limit which gives us \(a=b\in C\). This
shows that \(C\) is closed.

Now assume that \(C\) is unbounded.
Then for all \(n\in\mathbb{N}\), there
exists some \(a_n\in C\) with \(|a_n|\geq n\). Consider an arbitrary
subsequence \((a_{n_k})_{k\in\mathbb{N}}\). Due to \(|a_{n_k}|\geq n_k\geq k\), we have that
\((a_{n_k})_{k\in\mathbb{N}}\) is
unbounded, i.e., it cannot be convergent. This is also a contradiction
to compactness.

“(ii)\(\Rightarrow\)(i)”: Let \(C\) be closed and bounded. Let \((a_n)_{n\in\mathbb{N}}\) be a sequence in
\(C\). The boundedness of \(C\) then implies the boundedness of \((a_n)_{n\in\mathbb{N}}\). By the Theorem of
Bolzano-Weierstraß, there exists a convergent subsequence \((a_{n_k})_{k\in\mathbb{N}}\), i.e., \[\lim_{k\to\infty}a_{n_k}=a\] for some
\(a\in\mathbb{K}\). For compactness, we
now have to show that \(a\in C\).
However, this is guaranteed by the closedness of \(C\).\(\Box\)

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