Proving discontinuity – "Math for Non-Geeks"

verweis=Math for Non-Geeks

Overview

Recall that whenever we want to prove the negation of a statement about some elements of a set, we need to show that there as at least one element in that set for which the statement is not true. So, in order to prove the discontinuity of a function, all you have to show is that the function has (at least) one point of discontinuity. There are several methods available for proving the existence of a point of discontinuity:

Sequence criterion: Show that the function doesn't fulfill the sequence criterion at one particular point
Considering the left- and right-sided limits: Calculate the left-sided and right-sided limits of the function at a particular point. If either one of these limits doesn't exist, or if the limits are different, then the function is discontinuous at that point.
Epsilon-Delta Criterion: Show that the function doesn't fulfill the epsilon-delta criterion at a particular point.

Sequence Criterion

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Review: Sequence Criterion

Definition (Sequence criterion of continuity for a single argument)

A function $f : D \to ℝ$ with $D \subseteq ℝ$ is continuous at an argument $x_{0} \in D$ , if for all squences $(x_{n})_{n \in ℕ}$ with $\forall n \in ℕ : x_{n} \in D$ and $lim_{n \to \infty} x_{n} = x_{0}$ there is:

$lim_{n \to \infty} f (x_{n}) = f (lim_{n \to \infty} x_{n}) = f (x_{0})$

Sketch of the proof

In order to show that a function $f : D \to ℝ$ is discontinuous at $x_{0} \in D$ using the sequence criterion, we need to find one specific sequence of arguments $(x_{n})_{n \in ℕ}$ with $x_{n} \in D$ for all $n \in ℕ$ which is converging to $x_{0}$ , such that the sequence of function values ${(f (x_{n}))}_{n \in ℕ}$ does not converge to $f (x_{0})$ . So there shall be $lim_{n \to \infty} x_{n} = x_{0}$ but $lim_{n \to \infty} f (x_{n}) \neq f (x_{0})$ . In order for $lim_{n \to \infty} f (x_{n}) \neq f (x_{0})$ to hold, there are two cases to be distinguished:

The sequence of function values ${(f (x_{n}))}_{n \in ℕ}$ diverges.
The sequence of function values ${(f (x_{n}))}_{n \in ℕ}$ converges, but its limit is not $f (x_{0})$ .

Therefore, a proof of discontinuity using the sequence criterion could take the following form:

Let

f : \dots

be a function defined by

f (x) = \dots

. This function is discontinuous at

x_{0} = \dots

for the following reason: We take the sequence

(x_{n})_{n \in ℕ}

with

x_{n} = \dots

and all these elements are inside the domain of definition for

f

. This sequence converges to

$lim_{n \to \infty} x_{n} = \dots = x_{0}$

However, there is $lim_{n \to \infty} f (x_{n}) \neq f (x_{0})$ . And instead, there is

To do:
Finish this part.

...Proof that ${(f (x_{n}))}_{n \in ℕ}$ diverges or that the limit of ${(f (x_{n}))}_{n \in ℕ}$ exists and is different from $f (x_{0})$ ...

Example exercises

Exercise (Discontinuity of the topological sine function)

Prove discontinuity of the following function:

$f : ℝ \to ℝ : x \mapsto {\begin{matrix} \sin (\frac{1}{x}) & x \neq 0 \\ 0 & x = 0 \end{matrix}$

How to get to the proof? (Discontinuity of the topological sine function)

Discontinuity of $f$ means that this function has at least one argument where $f$ is discontinuous. For each $x \neq 0$ , $f$ is equal to the function $\sin (\frac{1}{x})$ in a sufficiently small neighbourhood of $x$ . Since $\sin (\frac{1}{x})$ is just a composition of continuous functions, it is continuous itself and therefore, $f$ must be continuous for all $x \neq 0$ , as well. So we know that the discontinuity may only be situated at $x = 0$ .

In order to prove that $f$ is discontinuous at $x = 0$ , we need to find a sequence of arguments $(x_{n})_{n \in ℕ}$ converging to $lim_{n \to \infty} x_{n} = 0$ but with $lim_{n \to \infty} f (x_{n}) \neq f (0)$ . To find such a function, let us take a look at the graph of the function $f$ :

In this figure, we recognize that $f$ takes any value between $- 1$ and $1$ infinitely often in the vicinity of $x = 0$ . So, for instance, we may just choose $(x_{n})_{n \in ℕ}$ such that $f (x_{n})$ is always $1$ . This guarantees that $lim_{n \to \infty} f (x_{n}) = 1 \neq 0 = f (0)$ - and actually any other real number $f (x_{n}) \neq 0$ between $- 1$ and $1$ in place of $f (x_{n}) = 1$ would do the job. But we need to make a specific choice for $x_{n}$ , and $f (x_{n}) = 1$ is a very simple one. In addition, we will choose $x_{n}$ to converges to zero from above.

The following figure also contains the sequence elements $x_{n}$ beside our function $f$ . We may clearly see that for $x_{n} \to 0$ the sequence of function values converges to $1$ , which is different from the function value $f (0) = 0$ :

But what are the exact values of these $x_{n}$ for which we would like to have $f (x_{n}) = 1$ To answer this question, let us resolve the equation $f (x) = 1$ for $x$ :

$\begin{matrix} \begin{matrix} f (x) & = 1 \\ \overset{f (0) \neq 0}{⟺} & \sin (\frac{1}{x}) & = 1 \\ ⟺ & \exists k \in ℤ : & \frac{1}{x} & = \frac{π}{2} + 2 k π \\ ⟺ & \exists k \in ℤ : & x & = \frac{1}{\frac{π}{2} + 2 k π} \end{matrix} \end{matrix}$

So for each $x = \frac{1}{\frac{π}{2} + 2 k π}$ with $k \in ℤ$ , we have $f (x) = 1$ . In order to get positive $x_{n}$ converging to zero from above, we may for instance choose $x_{n} = \frac{1}{\frac{π}{2} + 2 n π}$ . In that case:

$lim_{n \to \infty} x_{n} = lim_{n \to \infty} \frac{1}{\frac{π}{2} + 2 n π} = 0$

And we have seen that $lim_{n \to \infty} f (x_{n}) = 1 \neq 0 = f (0)$ . So we found just a sequence of arguments $(x_{n})_{n \in ℕ}$ , which proves discontinuity of $f$ at $x = 0$ .

Proof (Discontinuity of the topological sine function)

Let $f : ℝ \to ℝ$ with $f (x) = \sin (\frac{1}{x})$ for $x \neq 0$ and $f (0) = 0$ . We consider the sequence $(x_{n})_{n \in ℕ}$ defined by $x_{n} = \frac{1}{\frac{π}{2} + 2 n π}$ . For this sequence:

$lim_{n \to \infty} x_{n} = lim_{n \to \infty} \frac{1}{\frac{π}{2} + 2 n π} = 0$

And there is:

$\begin{matrix} lim_{n \to \infty} f (x_{n}) & = lim_{n \to \infty} f (\frac{1}{\frac{π}{2} + 2 n π}) \\ = lim_{n \to \infty} \sin (\frac{1}{\frac{1}{\frac{π}{2} + 2 n π}}) \\ = lim_{n \to \infty} \sin (\frac{π}{2} + 2 n π) \\ = lim_{n \to \infty} 1 = 1 \end{matrix}$

Hence, $lim_{n \to \infty} f (x_{n}) = 1 \neq 0 = f (0)$ although $lim_{n \to \infty} x_{n} = 0$ . This proves that $f$ is discontinuous at $x = 0$ and therefore it is a discontinuous function.

Epsilon-Delta Criterion

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Review: Epsilon-Delta Criterion

Definition (Epsilon-Delta definition of discontinuity)

A function $f : D \to ℝ$ with $D \subseteq ℝ$ is discontinuous at $x_{0} \in D$ , if and only if there is an $ϵ > 0$ , such that for all $δ > 0$ a $x \in D$ with $| x - x_{0} | < δ$ and $| f (x) - f (x_{0}) | \geq ϵ$ exists. Mathematically written, $f$ is discontinuous at $x_{0} \in D$ iff

$\exists ϵ > 0 \forall δ > 0 \exists x \in D : | x - x_{0} | < δ \land | f (x) - f (x_{0}) | \geq ϵ$

General proof structure

The Epsilon-Delta criterion of discontinuity can be formulated in predicate logic as follows:

$\exists ϵ > 0 \forall δ > 0 \exists x \in D : | x - x_{0} | < δ \land | f (x) - f (x_{0}) | \geq ϵ$

From here we get a scheme that allows us to prove the discontinuity of a function using the delta-epsilon criterion:

$\begin{matrix} \underset{\exists ϵ > 0}{\underset{⏟}{Choose ϵ = \dots}} \underset{\forall δ > 0}{\underset{⏟}{Let δ > 0 arbitrary.}} \\ \underset{\exists x \in D}{\underset{⏟}{Choose x = \dots We have x \in D, since \dots}} \\ We have: \\ Proof for | x - x_{0} | < δ \\ Proof for | f (x) - f (x_{0}) | \geq ϵ \end{matrix}$

Example exercise

Exercise (Discontinuity of the topological sine fucntion)

Prove the discontinuity at $x_{0} = 0$ for the topological sine function:

$f : ℝ \to ℝ : x \mapsto {\begin{matrix} \sin (\frac{1}{x}) & x \neq 0 \\ 0 & x = 0 \end{matrix}$

How to get to the proof? (Discontinuity of the topological sine fucntion)

In this exercise, discontinuity has to be shown for a given function. This is done by the negation of the epsilon-delta criterion. Our objective is to find both an $ϵ > 0$ and an $x \in ℝ$ , such that $| x - x_{0} | < δ$ and $| f (x) - f (x_{0}) | \geq ϵ$ . Here, $x$ may be chosen depending on $δ$ , while $ϵ$ has to be the same for all $δ > 0$ . For a solution, we may proceed as follows

Step 1: Simplify the target inequality

First, we may simplify the two inequalities which have to be fulfilled by plugging in $x_{0} = 0$ and $f (x_{0}) = 0$ . We therefore get: $| x | < δ$ and $| f (x) | \geq ϵ$ .

Step 2: Choose a suitable $ϵ > 0$

We consider the graph of the function $f$ . It will help us finding the building bricks for our proof:

We need to find an $ϵ > 0$ , such that there are arguments in each arbitrarily narrow interval $(x_{0} - δ, x_{0} + δ) = (- δ, δ)$ whose function values have a distance larger than $ϵ$ from $f (x_{0}) = f (0) = 0$ . Visually, no matter how small the width $δ$ of the $2 ϵ$ - $2 δ$ -rectangle is chosen, there will always be some points below or above it.

Taking a look at the graph, we see that our function oscillates between $- 1$ and $1$ . Hence, $ϵ \leq 1$ may be useful. In that case, there will be function values with $| f (x) | \geq 1$ in every arbitrarily small neighborhood around $x_{0} = 0$ . We choose $ϵ = \frac{1}{2}$ . This is visualized in the following figure:

After $ϵ$ has been chosen, an arbitrary $δ > 0$ will be assumed, for which we need to find a suitable $x$ . This is what we will do next.

Step 3: Choice of a suitable $x \in ℝ$

We just set $ϵ = \frac{1}{2}$ . Therefore, $| \sin (\frac{1}{x}) | \geq \frac{1}{2}$ has to hold. So it would be nice to choose an $x$ with $\sin (\frac{1}{x}) = 1$ . Now, $\sin (a) = 1$ is obtained for any $a = \frac{π}{2} + 2 k π$ with $k \in ℤ$ . The condition for $x$ such that the function gets 1 is therefore:

$\frac{1}{x} = \frac{π}{2} + 2 k π ⟺ x = \frac{1}{\frac{π}{2} + 2 k π}$

So we found several $x$ , where $| f (x) | \geq ϵ$ . Now we only need to select one among them, which satisfies $| x | < δ$ for the given $δ$ . Our $x$ depend on $k$ . So we have to select a suitable $k \in ℤ$ , where $| x | < δ$ . To do so, let us plug $x = \frac{1}{\frac{π}{2} + 2 k π}$ into this inequality and solve it for $k$ :

$\begin{matrix} | x | < δ & ⟺ | \frac{1}{\frac{π}{2} + 2 k π} | < δ \\ ⟺ \frac{1}{\frac{π}{2} + 2 k π} < δ \\ ⟺ 2 k π + \frac{π}{2} > \frac{1}{δ} \\ ⟺ 2 k π > \frac{1}{δ} - \frac{π}{2} \\ ⟺ k > \frac{1}{2 π} (\frac{1}{δ} - \frac{π}{2}) \end{matrix}$

So the condition on $k$ is $k > \frac{1}{2 π} (\frac{1}{δ} - \frac{π}{2})$ . If we choose just any natural number $k$ above this threshold $\frac{1}{2 π} (\frac{1}{δ} - \frac{π}{2})$ , then $| x | < δ$ will be fulfilled. Such a $k$ has to exist by Archimedes' axiom (for instance by flooring up the right-hand expression). So let us choose such a $k$ and define $x$ via $x = \frac{1}{\frac{π}{2} + 2 k π}$ . This gives us both $| x | < δ$ and $| f (x) | \geq ϵ$ . So we got all building bricks together, which we will now assemble to a final proof:

Proof (Discontinuity of the topological sine fucntion)

Choose $ϵ = \frac{1}{2}$ and let $δ > 0$ be arbitrary. Choose a natural number $k$ with $k > \frac{1}{2 π} (\frac{1}{δ} - \frac{π}{2})$ . Such a natural number $k$ has to exist by Archimedes' axiom. Further, let $x = \frac{1}{\frac{π}{2} + 2 k π}$ . Then:

$\begin{matrix} k > \frac{1}{2 π} (\frac{1}{δ} - \frac{π}{2}) & ⟹ 2 k π > \frac{1}{δ} - \frac{π}{2} \\ ⟹ 2 k π + \frac{π}{2} > \frac{1}{δ} \\ ↓ \frac{1}{a} > \frac{1}{b} > 0 ⟺ 0 < a < b \\ ⟹ \frac{1}{\frac{π}{2} + 2 k π} < δ \\ ↓ \frac{1}{\frac{π}{2} + 2 k π} > 0 \\ ⟹ | \frac{1}{\frac{π}{2} + 2 k π} | < δ \\ ↓ x = \frac{1}{\frac{π}{2} + 2 k π} \\ ⟹ | x | < δ \end{matrix}$

In addition:

$\begin{matrix} | f (x) - f (0) | & = | \sin (\frac{1}{x}) - 0 | \\ ↓ x = \frac{1}{\frac{π}{2} + 2 k π} \\ = | \sin (\frac{π}{2} + 2 k π) | \\ ↓ \sin (\frac{π}{2} + 2 k π) = 1 \\ = | 1 | \geq \frac{1}{2} = ϵ \end{matrix}$

Hence, the function is discontinuous at $x_{0} = 0$ .

Exercises

Epsilon-delta criterion: Sign function

Composition of continuous functions →

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