For some function \(f: [a, b] \rightarrow \mathbb{R}\) continuous, the Extreme Value Theorem and Intermediate Value Theorem apply. More generally, for \(f: (X, d_x) \rightarrow (Y, d_y)\) continuous, we get

  • \(k \subset X\) is compact \(\implies f(k) \subset Y\) is compact.
  • \(E \subset X\) is connected \(\implies f(E)\) is connected.

Recall that the Heine-Borel Theorem gives us \(k \subset \mathbb{R}^n \textrm{ compact } \iff K \textrm{ is closed and bounded}\).

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Def: \((X,d)\) metric space:

\((X_n) \in X\) seq. Say \((X_n)\) converges to \(x_0 \in X\).If \(\forall \epsilon > 0, \exists N > 0\) such that \(d(x_n, x_0) < \epsilon \forall n> N\).

Def: \(f: (X, d_x) \rightarrow (Y, d_y)\)

We say \(f\) is continuous at \(x_0 \in X\) if \(\forall\) sequence \((x_n) \in X\) that converges to \(x_0\), we have \((f(x_n)) \in Y\) converges to \(f(x_0)\).

Continuity For Functions

Theorem: \(f: X \rightarrow Y\) is continuous at \(x_0\) \(\iff\) \(\forall \epsilon > 0, \exists \delta > 0\) such that \(d_x(x, x_0) < \delta \implies \delta_y(f(x), f(x_0)) < \epsilon\).

Proof: (←): Given the above premise, let \((x_n) \in X\) be a sequence such that \(\lim x_n = x_0\). We want to prove \(\lim f(x_n) \leq f(x_0)\). In other words, \(\forall \epsilon > 0, \exists N > 0\) such that \(d_y(f(x_n)-f(x_0)) < \epsilon\ \forall n > N\)

\(\exists \delta > 0\) such that \(d_x(x, x_0) < \delta \implies d_y(f(x), f(x_0)) < \epsilon\) by the premise.
\(\exists N > 0\) such that \(d_x(x, x_0) < \delta\) for any \(n>N\) since \(\lim x_n = x_0\)

At this point, \(N\) depends on \(\delta\) and \(\delta\) depends on \(\epsilon\). \(\forall \epsilon > 0\), we found \(N > 0\) such that if \(n>N\), then \(d_x(x_n, x_0) < \delta\), meaning \(d_y(f(x_n), f(x_0)) < \epsilon\).

(→): Proof by Contradiction.

The converse (ie. assuming the conclusion is false) is \(\exists \epsilon > 0\) such that \(\forall \delta > 0\) such that \(\forall \delta > 0\), \(\exists x \in X\) such that \(d(x, x_0) < \delta\) but \(d(f(x), f(x_0)) \geq \epsilon\)

Let \(\delta = 1\). \(\exists x_n \in X\) such that \(d(x_1, x_0) < 1\) but \(d(f(x_1), f(x_0)) \geq \epsilon\)
Let \(\delta = 1/n\). \(\exists x_n \in X\) such that \(d(x_1, x_0) < 1/n\) but \(d(f(x_1), f(x_0)) \geq \epsilon\)

Consider \((x_n)\). We have \(\lim x_n = x_0\) but \(f(x_n)\) doesn’t converge to \(f(x_0)\).

The converse: \(f: S \rightarrow \mathbb{R}, S \in \mathbb{R}\). \(f\) is continuous at \(x_0 \in S\) if and only if \(\forall \epsilon > 0, \exists \delta > 0\) such that \(\vert x - x_0 \vert < \delta\) implies that \(\vert f(x) - f(x_0) \vert < \epsilon\).

Example 1

Prove that \(f\) is not continous at \(x=2\).

\[f(x) = \begin{cases} x, & x \neq 2\\ 3, & x=2\\ \end{cases}\]

Intuition: If we take an delta-neighborhood around \(x=2\), we will see there exists no points around it in the image.

Proof: Take \(\epsilon = 1/2\). \(\forall \delta > 0, \exists x \in \mathbb{R}\) such that \(\vert x - 2 \vert < \delta\), but \(\vert f(x) - 3 \vert > 1/2\).

Example 2

Let \(f(x) = x\). We want to show that it is continuous everywhere.

Proof: \(\forall x_0 \in \mathbb{R}\), \(\forall \epsilon > 0\), take \(\delta = \epsilon\). Then, \(\vert x-x_0 \vert < \delta = \epsilon \implies \vert f(x) - f(x_0) \vert < \epsilon\)

Example 3

Let \(f(x) = x^2\). We want to show that it is continuous everywhere.

Proof: \(\forall x_0 \in \mathbb{R}\), \(\forall \epsilon > 0\), we want to find \(\delta > 0\) such that if \(\vert x - x_0 \vert < \delta\), then \(\vert x^2 - x_0^2 \vert < \epsilon\).
Notice that \(\vert x^2 - x_0^2 \vert = \vert x - x_0 \vert \vert x + x_0 \vert\).

Take \(\delta = \min(\frac{\epsilon}{y \vert x_0 \vert}, \vert x_0 \vert)\). We get \(\vert x - x_0 \vert < \delta \leq \vert x_0 \vert\), and \(\vert x + x_0 \leq \vert x-x_0 \vert + 2 \vert x_0 \vert \leq 3\vert x_0 \vert\).
Then, \(\vert x - x_0 \vert \vert x + x_0 \vert\) \(< \delta * 3 \vert x_0 \vert \leq \frac{\epsilon}{4 \vert x_0} * 3 \vert x_0 \vert\) \(< \epsilon\).

Alternatively, take \(\delta = \min\{1, \frac{\epsilon}{1 + 2\vert x_0 \vert} \}\). We get \(\vert x - x_0 \vert < \delta \leq 1\), and \(\vert x + x_0 \vert \leq \vert x - x_0 \vert + 2 \vert x_0 \vert < 1 + 2 \vert x_0 \vert\). This gives us \(\vert x - x_0 \vert \vert x + x_0 \vert < \delta * (1 + 2\vert x_0 \vert)\leq \epsilon\).


\(\delta > 0\) typically depends on both \(\epsilon > 0\) and \(x_0 \in X\).

Continuity For General Spaces

Theorem: \(f: (X, d_x) \rightarrow (Y, d_y)\) is continuous if and only if \(\forall U \subset Y\) open, \(f^{-1} (U) \subset X\) is open. Remark. \(f^{-1}\) is not a function; it is defined as \(f^{-1} := {x \in X \vert f(x) \in U}\).


(→) \(f\) continuous, \(U \subset Y\) open. We want to show that \(f^{-1}(U) \subset X\) is open, ie. \(\forall x \in f^{-1}(U), \exists r > 0\) such that \(B_r(x) \subset f^{-1}(U)\).
Since \(U\) is open, \(\exists \epsilon > 0\) such that \(B_\epsilon(y) \subset U\).
Since \(f\) is continuous, we can find a \(\delta\) such that if \(z \in B_\delta(x)\), then \(f(z) \in B_\epsilon(y)\), which implies that \(z \in f^{-1}(U)\), and therefore \(B_\delta(x) \subset f^{-1}(U)\).

(←) Given that \(\forall U \subset Y\) open, \(f^{-1} (U) \subset X\) is open, \(\forall x_0 \in X, \forall \epsilon > 0\), we want to show that \(\exists \delta > 0\) such that \(d(x, x_0) < \delta \implies d(f(x), f(x_0)) < \epsilon\).
\(B_{\epsilon}(f(x_0)) \subset Y\) is open. By our premise, \(f^{-1}(B_{\epsilon}(f(x_0))) \subset X\) is open.
Therefore, \(\exists \delta > 0\) such that \(B_\delta(x_0) \subset f^{-1} (B_\epsilon (f(x_0)))\), hence if \(d(x, x_0)< \delta\), then \(d(f(x), f(x_0)) < \epsilon\).