Three Faces of Continuity

Continuity has three faces: the

\varepsilon

\delta

argument, the sequential test, and preimages of open sets. Placing

f(x) = x^2

next to a jump function shows the three faces saying the same thing.

Examples & Exercises

Prove completely that $\varepsilon$ - $\delta$ continuity of $f$ at $x_0$ implies sequential continuity: $f(x_n) \to f(x_0)$ for every sequence $x_n \to x_0$ .

\forall\varepsilon>0\ \exists\delta>0:\ \lvert x-x_0\rvert<\delta \Rightarrow \lvert f(x)-f(x_0)\rvert<\varepsilon

x_n \to x_0 \ \Longrightarrow\ f(x_n) \to f(x_0)

f(x) = x^2

\varepsilon = 0.3

\delta_{\max} \approx 0.54

(

=\sqrt{0.3}

) keeps every

x

inside the band — this

\delta

is the heart of the proof below.

Show proof

Assume $f$ is $\varepsilon$ - $\delta$ continuous at $x_0$ : for every $\varepsilon > 0$ there is $\delta > 0$ with $|x - x_0| < \delta \Rightarrow |f(x) - f(x_0)| < \varepsilon$ .
Take a sequence $x_n \to x_0$ and $\varepsilon > 0$ . Fetch the $\delta$ for this $\varepsilon$ from above; since $x_n \to x_0$ , there is $N$ with $n > N \Rightarrow |x_n - x_0| < \delta$ .
For all $n > N$ , $|f(x_n) - f(x_0)| < \varepsilon$ , so $f(x_n) \to f(x_0)$ . Since $\varepsilon > 0$ was arbitrary, sequential continuity holds.

Prove by contrapositive: if $f$ is not $\varepsilon$ - $\delta$ continuous at $x_0$ , one can build a sequence $(x_n)$ with $x_n \to x_0$ but $f(x_n) \not\to f(x_0)$ .

\exists\varepsilon_0>0\ \forall\delta>0\ \exists x:\ \lvert x-x_0\rvert<\delta \text{ and } \lvert f(x)-f(x_0)\rvert \ge \varepsilon_0

\lvert x_n - x_0\rvert < \tfrac{1}{n}, \qquad \lvert f(x_n) - f(x_0)\rvert \ge \varepsilon_0

The jump function,

\varepsilon = 0.3

x_n = (-1)^n/n \to 0

, yet

f(x_n)

bounces near

0

and

1

and never settles at

f(0) = 1

— the proof below builds this sequence in general form.

Show proof

Assume $f$ is not $\varepsilon$ - $\delta$ continuous at $x_0$ . Negating the definition: some $\varepsilon_0 > 0$ exists such that for every $\delta > 0$ there is $x$ with $|x - x_0| < \delta$ and $|f(x) - f(x_0)| \ge \varepsilon_0$ .
For each natural $n$ , plug in $\delta = 1/n$ and pick $x_n$ : $|x_n - x_0| < 1/n$ and $|f(x_n) - f(x_0)| \ge \varepsilon_0$ .
By the squeeze theorem, $|x_n - x_0| < 1/n \to 0$ , so $x_n \to x_0$ . But $|f(x_n) - f(x_0)| \ge \varepsilon_0$ for every $n$ , so $f(x_n) \not\to f(x_0)$ — if $f(x_n) \to f(x_0)$ there would have to be $n$ with $|f(x_n) - f(x_0)| < \varepsilon_0/2$ , a contradiction. Hence sequential continuity fails.

Prove completely in both directions that $f$ is continuous on its whole domain if and only if the preimage $f^{-1}(V)$ of every open set is open.

f \text{ continuous} \iff f^{-1}(V) \text{ open for every open } V

V = (f(x_0)-\varepsilon,\ f(x_0)+\varepsilon), \qquad U = f^{-1}(V) \ni x_0 \text{ open}

The jump function,

\varepsilon = 0.3

: the preimage of

V = (0.7, 1.3)

contains no interval around

0

(on the left only

|x| \approx 0.84

and beyond qualifies) — the preimage of an open set is not open.

Show proof

( $\Rightarrow$ ) Let $f$ be continuous and $V$ open. Take $x_0 \in f^{-1}(V)$ ; then $f(x_0) \in V$ , and since $V$ is open there is $\varepsilon > 0$ with $(f(x_0)-\varepsilon, f(x_0)+\varepsilon) \subset V$ .
Continuity at $x_0$ gives $\delta > 0$ with $|x - x_0| < \delta \Rightarrow |f(x) - f(x_0)| < \varepsilon$ . Then every $x$ in $(x_0-\delta, x_0+\delta)$ satisfies $f(x) \in (f(x_0)-\varepsilon, f(x_0)+\varepsilon) \subset V$ , so $x \in f^{-1}(V)$ . Thus $x_0$ is an interior point of $f^{-1}(V)$ ; $x_0$ was arbitrary, so $f^{-1}(V)$ is open.
( $\Leftarrow$ ) Assume the preimage of every open set is open. For $x_0$ and $\varepsilon > 0$ , $V = (f(x_0)-\varepsilon, f(x_0)+\varepsilon)$ is open, so $U = f^{-1}(V)$ is open with $x_0 \in U$ . Hence some $\delta > 0$ has $(x_0-\delta, x_0+\delta) \subset U$ , and $|x - x_0| < \delta$ gives $f(x) \in V$ , i.e. $|f(x) - f(x_0)| < \varepsilon$ . Since $x_0$ was arbitrary, $f$ is continuous.