Compactness
Subsequences
Definition 1. Let $a \colon \N \to X$ be a sequence, and $n \colon \N \to \N$ be a strictly increasing sequence. We say $a \circ n$ is a subsequence of $a$.
Remark 2. We usually denote the sequence by $(a_n)_{n \in \N}$, or just $(a_n)$ implicitly assuming $n \in \N$. We usually denote subsequences by $(a_{n_k})_{k \in \N}$, or just $(a_{n_k})$, implicitly assuming $k \in \N$. (Here $(n_k)_{k \in \N}$ is a strictly increasing sequence in $\N$.)
Example 3. If $(a_n)$ is a sequence, then $(a_{2n +1})$ is the subsequence of odd terms, and $(a_{2n})$ is the subsequence of even terms.
Proposition 4. If $a_n \to \ell$ as $n \to \infty$, then for any subsequence $(a_{n_k})$ we must also have $a_{n_k} \to \ell$ as $k \to \infty$.
Lemma 5 (Rising sun). Any sequence in $\R$ has a monotone subsequence.
Proof sketch. Intuition: Imagine the sequence as buildings of height $a_n$ built at mile marker $n$ on a straight road. You’re trying to see the sunrise from the top of a building. Every time you can, you try and go closer. If you can always do this, you have a decreasing subsequence; if at some point you can’t you have a strictly increasing subsequence.
Rigorous version: Let $G = \set{n \st a_n \geq a_m \forall m \geq n }$. If $G$ is infinite, ordering the elements of $G$ will give you a decreasing subsequence. If $G$ is finite (or empty), show that there exists a strictly increasing sequence of indexes $(n_k)_{k \in \N}$ with $n_1 = \min \N - G$ such that $(a_{n_k})$ is also strictly increasing.
Sequential compactness
Definition 6. A set $K \subseteq X$ is called sequentially compact if any sequence in $K$ as a convergent subsequence in $K$. Explicitly, if $(a_n)$ is a sequence with $a_n \in K$ for all $n \in \N$, then there exists a subsequence $(a_{n_k})$ such that $\lim_{k \to \infty} a_{n_k} \in K$.
Example 7. Any finite set is sequentially compact.
Example 8. $\R$ is not sequentially compact.
Proposition 9. Any sequentially compact set is closed and bounded.
Proposition 10. In a general metric space, closed and bounded sets need not be sequentially compact.
Theorem 11 (Bolzano–Weierstrass). Any closed and bounded set in $\R^d$ is sequentially compact.
Remark 12. In arbitrary metric spaces, the generalization of Bolzano Weierstrass theorem asserts that a set is sequentially compact if and only if it is totally bounded and complete.
Continuity and Compactness
Extreme value theorem
Theorem 13. If $K \subseteq X$ is sequentially compact, and $f \colon X \to Y$ is continuous, then $f(K) \subseteq Y$ is sequentially compact.
Remark 14. If $K \subseteq Y$ is sequentially compact, and $f \colon X \to Y$ is continuous, then $f^{-1}(K) \subseteq X$ need not be sequentially compact.
Corollary 15. If $K$ is compact, and $f \colon K \to \R$ is continuous, then $f$ is bounded and there exists $m, M \in K$ such that \begin{equation} f(m) = \inf_{x \in K} f(x) \quad\text{and}\quad f(M) = \sup_{x \in K} f(x) \,. \end{equation}
Corollary 16 (Extreme value theorem). If $f \colon [a, b] \to \R$ is continuous, then $f$ is bounded and attains its bounds.
Continuity of inverses
Remark 17. If $f \colon X \to Y$ is continuous and bijective, then $f^{-1}\colon Y \to X$ need not be continuous.
Theorem 18. If $X$ is sequentially compact and $f \colon X \to Y$ is continuous and bijective, then $f^{-1}\colon Y \to X$ is also continuous.
Proof sketch. Option 1: Check that for every closed set $C \subseteq X$, $f(C) \subseteq Y$ is closed.
Option 2: Suppose for contradiction you had $y \in Y$ and a sequence $y_n \xrightarrow{n \to \infty} y$, but $f^{-1}(y_n) \not\to f^{-1}(y)$ as $n \to \infty$. Use compactness of $X$ and continuity of $f$ to find a contradiction.
Uniform continuity
Definition 19. We say $f \colon X \to Y$ is uniformly continuous if for every $\epsilon > 0$ there exists $\delta > 0$ such that for every $x, y \in X$, we have $d(x, y) < \delta$ implies $d(f(x), f(y)) < \epsilon$.
Proposition 20. Any uniformly continuous function is continuous.
Proposition 21. In general, any continuous function need not be uniformly continuous. (Example $f(x) = x^2$, with $X = Y = \R$.)
Theorem 22. If $X$ is sequentially compact then any continuous function on $X$ is also uniformly continuous.
Proof sketch. Suppose for contradiction $f$ is not uniformly continuous. Then show there exists some $\epsilon > 0$ and sequences $(x_n), (y_n) \in X$ such that $d(x_n, y_n) < 1/n$ and $d(f(x_n), f(y_n)) > \epsilon$. Pass to a convergent subsequence and find a contradiction.
Continuous extensions
Suppose $D \subseteq X$, and $f \colon D \to Y$ is continuous. Does there exist a continuous function $\bar f \colon X \to Y$ such $\bar f(x) = f(x)$ for every $x \in D$? Such a function is called a continuous extension of $f$.
In general, the answer is no. For example pick $D = (0, 1)$, $X = [0, 1]$ and $f(x) = 1/x$. However, if $D$ is dense, and $f$ is uniformly continuous then there exists a unique extension $\bar f$.
Theorem 23. If $D \subseteq X$ is dense, $Y$ is complete, and $f \colon D \to Y$ is continuous, then there is a unique continuous function $\bar f \colon X \to Y$ such that $\bar f(x) = f(x)$ for every $x \in D$.