Lecture 25 - Continuity

Last time we talked about the definition of a limit. Today we will talk about continuity.

continuity, epsilon-delta definition

Let $f : A \to R$ where $A \subseteq R$ and $c \in A$ . Then $f$ is continuous at $c$ if $\forall ε > 0 \exists δ > 0 \forall x \in A (| x - c | < δ \to | f (x) - f (c) | < ε)$

Compare and contrast this with the definition of a Functional Limit.

Here are some remarks:

If $c$ is Isolated Point then $f$ is automatically continuous at $c$ .
If $c$ is a Limit Point of $A$ then this can be phrased in terms of limits.

These are bundled up into a nice little theorem.

Characterization of Continuity

If $f : A \to R$ where $A \subseteq R$ and $c \in A$ , then the following are equivalent:

$f$ is continuous at $c$ .
$\forall ε > 0 \exists δ > 0 \forall x \in A (x \in A \cap V_{ε} (c) \to f (x) \in V_{ε} (f (c)))$
$\forall (x_{n}) \subseteq A$ where $x_{n} \to c$ , then $f (x_{n}) \to f (c)$ .
If specifically $c$ is a Limit Point of $A$ , then each of these is equivalent to:
$lim_{x \to c} f (x) = f (c)$

Proof

Notice the first one is just a rewording using Epsilon-Neighborhood notation. Thus $(1) ⟺ (2)$ .

#todo/school : Finish this proof 📅 2024-11-08 ✅ 2024-11-06

( $(3) ⟺ (1)$ ): Use an argument similar to the Sequential Criterion For Functional Limits, with some modifications for when $x_{n} = c$ .

( $(4) ⟺ (1)$ ): Consider the Functional Limit definition. Just observe the case that $x = c$ (which isn't included in the definition for Functional Limits) leads to the requirement that $f (c) \in V_{ε} (f (c))$ , which is trivially true.

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As a bit of a corollary, we can get a characterization for discontinuity.

Criterion for Discontinuity

Let $f : A \to R$ and let $c \in A$ be a limit point of $A$ . If there exists a sequence $(x_{n}) \subseteq A$ where $(x_{n}) \to c$ but such that $f (x_{n}) ↛ f (c)$ , we may conclude that $f$ is not continuous at $c$ .

Similar to how we can do operations on limits, we can similar do the same for continuity:

Algebraic Continuity Theorem

Let $f, g : A \to R$ where $A \subseteq R$ and $c \in A$ . Suppose $f, g$ each are continuous at $c$ . then:

$α f$ (for all $α \in R$ )
$f + g$
$f g$
Are continuous at $c$ . If further $g (c) \neq 0$ then $\frac{f}{g}$ is continuous at $c$ too.

Proof

Apply (3) from the Characterization of Continuity with the Algebraic Limit Theorem for Functional Limits for sequences

#todo/school : Add backlinks as well as divide the notes between files and do this proof once done 📅 2024-11-06 ✅ 2024-11-06

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Continuous Examples

$f (x) = x$ where $A = R$ . Here $f$ is continuous at every $c \in R$ . Let's use the definition of continuity to show this.

Proof

Let $ε > 0$ . Choose $δ = ε > 0$ . Let $x \in A$ be arbitrary. Then if we suppose that $| x - c | < δ$ then:

| f (x) - f (c) | = | x - c | < δ = ε \Rightarrow | f (x) - f (c) | < ε

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All polynomials are continuous over the real line ( $A = R$ ) at every point on that line $c \in R$ . This comes directly from applying (1) except using multiplications from the Algebraic Continuity Theorem and scaling them by constants.
All rational functions are continuous at every point they are defined. This also comes from using the Algebraic Continuity Theorem and applying the division rule.
$f (x) = \sqrt{x}$ over $A = [0, \infty)$ is continuous for all $c \in A$ . Let's use the proof, but first doing some scratch:

If we take $c \geq 0$ , what do we know about:

\begin{aligned} | f (x) - f (c) | & = | \sqrt{x} - \sqrt{c} | \\ = | \sqrt{x} - \sqrt{c} | \cdot \frac{| \sqrt{x} + \sqrt{c} |}{| \sqrt{x} + \sqrt{c}} \\ = \frac{| x - c |}{| \sqrt{x} + \sqrt{c} |} \\ = \frac{| x - c |}{\sqrt{x} + \sqrt{c}} \end{aligned}

We want $| f (x) - f (c) | < ε$ for any $| x - c | < δ$ . Then:

\begin{aligned} \frac{| x - c |}{\sqrt{x} + \sqrt{c}} & \leq \frac{| x - c |}{\sqrt{c}} < ε \end{aligned}

Thus we want $| x - c | < ε \sqrt{c}$ implying that $δ = ε \sqrt{c}$ . But we have issues if $c = 0$ , so let's handle that separately:

| f (x) - f (c) | = | \sqrt{x} - 0 | = \sqrt{x} = \sqrt{| x - 0 |}

So if we make $| x - 0 | < ε^{2} = δ$ then that evaluates the RHS to $< ε$ as we wanted.

Proof

Let $ε > 0$ . We have two cases:

( $c = 0$ ): Choose $δ = ε^{2}$ . Then if $| x - 0 | < δ = ε^{2}$ then:

\begin{aligned} | x - 0 | & < ε^{2} \\ \sqrt{| x - 0 |} & < ε \\ \sqrt{x} & < ε \\ | f (x) - f (c) | & < ε \end{aligned}

as desired.
2. ( $c \neq 0$ ): Choose $δ = ε \sqrt{c}$ . Then if $| x - c | < δ$ then:

\begin{aligned} | f (x) - f (c) | & = | \sqrt{x} - \sqrt{c} | \\ = | \frac{(\sqrt{x} - \sqrt{c}) (\sqrt{x} + \sqrt{c})}{\sqrt{x} + \sqrt{c}} | \\ = \frac{| x - c |}{\sqrt{x} + \sqrt{c}} \\ \leq \frac{| x - c |}{\sqrt{c}} \\ < \frac{δ}{\sqrt{c}} \\ = ε \end{aligned}

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Discontinuous Examples

Take $f (x) = {\begin{cases} \sin (\frac{1}{x}) & x \neq 0 \\ 0 & x = 0 \end{cases}$ . This one we talked about how the limit at 0 didn't exist. But because $c$ is a limit point of the domain $R$ then because the limits don't exist then that implies that it can't be continuous at 0.
See Dirichlet's Function.
Define $h (x)$ as:

h (x) = {\begin{cases} x & x \in Q \\ 0 & x \notin Q \end{cases}

Then $h$ is still discontinuous at any $c \neq 0$ . However, at $c = 0$ we recover continuity. This is because:

| h (x) - h (0) | = | h (x) | \leq | x | = | x - 0 | < δ

so setting $δ = ε$ works here.

Notice that we can write $h (x) = x g (x)$ where $g (x)$ is defined above. But

$h (x)$ is continuous at $c = 0$
$x$ is continuous at $c = 0$
But $g (x)$ is not continuous at $c = 0$ even though $g (x) = \frac{h (x)}{x}$ . The problem is, as always, that the limit is 0 for our denominator's function.