Three Faces of Continuity

This scene checks continuity in three different ways. The ε–δ argument, the sequential test, and preimages of open sets stand side by side.

The three faces say the same thing. Bands, sequences, or pulled-back open sets all give the same verdict. A smooth curve passes all three. A jump function fails all three at the same spot.

For example, look at one object in a 🪞 mirror from three sides. Front or side, it is the same object. Continuity is like that: many checks, one question.

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.