# Proving discontinuity – Serlo

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## 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 the 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

Main article: Sequential definition of continuity

### Review: Sequence Criterion

Definition (Sequence criterion of continuity for a single argument)

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

${\displaystyle \lim _{n\to \infty }f(x_{n})=f\left(\lim _{n\to \infty }x_{n}\right)=f(x_{0})}$

### Sketch of the proof

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

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

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

Let ${\displaystyle f:\ldots }$ be a function defined by ${\displaystyle f(x)=\ldots }$. This function is discontinuous at ${\displaystyle x_{0}=\ldots }$ for the following reason: We take the sequence ${\displaystyle (x_{n})_{n\in \mathbb {N} }}$ with ${\displaystyle x_{n}=\ldots }$ and all these elements are inside the domain of definition for ${\displaystyle f}$. This sequence converges to

${\displaystyle \lim _{n\to \infty }x_{n}=\ldots =x_{0}}$

However, there is ${\displaystyle \lim _{n\to \infty }f(x_{n})\neq f(x_{0})}$. And instead, there is ...Proof that ${\displaystyle \left(f(x_{n})\right)_{n\in \mathbb {N} }}$ diverges or that the limit of ${\displaystyle \left(f(x_{n})\right)_{n\in \mathbb {N} }}$ exists and is different from ${\displaystyle f(x_{0})}$ ...

### Example exercises

Exercise (Discontinuity of the topological sine function)

Prove discontinuity of the following function:

${\displaystyle f:\mathbb {R} \to \mathbb {R} :x\mapsto {\begin{cases}\sin \left({\frac {1}{x}}\right)&x\neq 0\\0&x=0\end{cases}}}$

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

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

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

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

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

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

{\displaystyle {\begin{aligned}{\begin{array}{rrrl}&&f(x)&=1\\[0.5em]{\overset {f(0)\neq 0}{\iff {}}}&&\sin \left({\frac {1}{x}}\right)&=1\\[0.5em]\iff {}&\exists k\in \mathbb {Z} :&{\frac {1}{x}}&={\frac {\pi }{2}}+2k\pi \\[0.5em]\iff {}&\exists k\in \mathbb {Z} :&x&={\frac {1}{{\frac {\pi }{2}}+2k\pi }}\end{array}}\end{aligned}}}

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

${\displaystyle \lim _{n\to \infty }x_{n}=\lim _{n\to \infty }{\frac {1}{{\frac {\pi }{2}}+2n\pi }}=0}$

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

Proof (Discontinuity of the topological sine function)

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

${\displaystyle \lim _{n\to \infty }x_{n}=\lim _{n\to \infty }{\frac {1}{{\frac {\pi }{2}}+2n\pi }}=0}$

And there is:

{\displaystyle {\begin{aligned}\lim _{n\to \infty }f(x_{n})&=\lim _{n\to \infty }f\left({\frac {1}{{\frac {\pi }{2}}+2n\pi }}\right)\\[0.5em]&=\lim _{n\to \infty }\sin \left({\frac {1}{\frac {1}{{\frac {\pi }{2}}+2n\pi }}}\right)\\[0.5em]&=\lim _{n\to \infty }\sin \left({\frac {\pi }{2}}+2n\pi \right)\\[0.5em]&=\lim _{n\to \infty }1=1\end{aligned}}}

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

## Epsilon-Delta Criterion

Main article: Epsilon-delta definition of continuity

### Review: Epsilon-Delta Criterion

Definition (Epsilon-Delta definition of discontinuity)

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

${\displaystyle \exists \epsilon >0\,\forall \delta >0\,\exists x\in D:|x-x_{0}|<\delta \land |f(x)-f(x_{0})|\geq \epsilon }$

### General proof structure

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

${\displaystyle {\color {Red}\exists \epsilon >0}\ {\color {RedOrange}\forall \delta >0}\ {\color {OliveGreen}\exists x\in D}:{\color {DarkOrchid}|x-x_{0}|<\delta }\land {\color {Blue}|f(x)-f(x_{0})|\geq \epsilon }}$

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

${\displaystyle {\begin{array}{l}{\color {Red}\underbrace {{\underset {}{}}{\text{Choose }}\epsilon =\ldots } _{\exists \epsilon >0}}\ {\color {RedOrange}\underbrace {{\underset {}{}}{\text{Let }}\delta >0{\text{ beliebig.}}} _{\forall \delta >0}}\\{\color {OliveGreen}\underbrace {{\underset {}{}}{\text{Choose }}x=\ldots {\text{ Es ist }}x\in D{\text{, weil}}\ldots } _{\exists x\in D}}\\{\color {Black}{\text{We have:}}}\\[0.5em]\quad \quad {\color {DarkOrchid}{\text{Proof for }}|x-x_{0}|<\delta }\\[0.5em]\quad \quad {\color {Blue}{\text{Proof for }}|f(x)-f(x_{0})|\geq \epsilon }\end{array}}}$

### Example exercise

Exercise (Discontinuity of the topological sine fucntion)

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

${\displaystyle f:\mathbb {R} \to \mathbb {R} :x\mapsto {\begin{cases}\sin \left({\frac {1}{x}}\right)&x\neq 0\\0&x=0\end{cases}}}$

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 ${\displaystyle \epsilon >0}$ and an ${\displaystyle x\in \mathbb {R} }$, such that ${\displaystyle |x-x_{0}|<\delta }$ and ${\displaystyle |f(x)-f(x_{0})|\geq \epsilon }$ . Here, ${\displaystyle x}$ may be chosen depending on ${\displaystyle \delta }$ , while ${\displaystyle \epsilon }$ has to be the same for all ${\displaystyle \delta >0}$ . 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 ${\displaystyle x_{0}=0}$ and ${\displaystyle f(x_{0})=0}$. We therefore get: ${\displaystyle |x|<\delta }$ and ${\displaystyle |f(x)|\geq \epsilon }$.

Step 2: Choose a suitable ${\displaystyle \epsilon >0}$

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

We need to find an ${\displaystyle \epsilon >0}$ , such that there are arguments in each arbitrarily narrow interval ${\displaystyle (x_{0}-\delta ,x_{0}+\delta )=(-\delta ,\delta )}$ whose function values have a distance larger than ${\displaystyle \epsilon }$ from ${\displaystyle f(x_{0})=f(0)=0}$ . Visually, no matter how small the width ${\displaystyle \delta }$ of the ${\displaystyle 2\epsilon }$-${\displaystyle 2\delta }$-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 ${\displaystyle -1}$ and ${\displaystyle 1}$ . Hence, ${\displaystyle \epsilon \leq 1}$ may be useful. In that case, there will be function values with ${\displaystyle |f(x)|\geq 1}$ in every arbitrarily small neighborhood around ${\displaystyle x_{0}=0}$. We choose ${\displaystyle \epsilon ={\tfrac {1}{2}}}$. This is visualized in the following figure:

After ${\displaystyle \epsilon }$ has been chosen, an arbitrary ${\displaystyle \delta >0}$ will be assumed, for which we need to find a suitable ${\displaystyle x}$. This is what we will do next.

Step 3: Choice of a suitable ${\displaystyle x\in \mathbb {R} }$

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

${\displaystyle {\frac {1}{x}}={\frac {\pi }{2}}+2k\pi \iff x={\frac {1}{{\frac {\pi }{2}}+2k\pi }}}$

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

{\displaystyle {\begin{aligned}\left|x\right|<\delta &\iff \left|{\frac {1}{{\frac {\pi }{2}}+2k\pi }}\right|<\delta \\[0.5em]&\iff {\frac {1}{{\frac {\pi }{2}}+2k\pi }}<\delta \\[0.5em]&\iff 2k\pi +{\frac {\pi }{2}}>{\frac {1}{\delta }}\\[0.5em]&\iff 2k\pi >{\frac {1}{\delta }}-{\frac {\pi }{2}}\\[0.5em]&\iff k>{\frac {1}{2\pi }}\left({\frac {1}{\delta }}-{\frac {\pi }{2}}\right)\end{aligned}}}

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

Proof (Discontinuity of the topological sine fucntion)

Choose ${\displaystyle \epsilon ={\tfrac {1}{2}}}$ and let ${\displaystyle \delta >0}$ be arbitrary. Choose a natural number ${\displaystyle k}$ with ${\displaystyle k>{\tfrac {1}{2\pi }}\left({\tfrac {1}{\delta }}-{\tfrac {\pi }{2}}\right)}$. Such a natural number ${\displaystyle k}$ has to exist by Archimedes' axiom. Further, let ${\displaystyle x={\tfrac {1}{{\frac {\pi }{2}}+2k\pi }}}$. Then:

{\displaystyle {\begin{aligned}k>{\frac {1}{2\pi }}\left({\frac {1}{\delta }}-{\frac {\pi }{2}}\right)&\implies 2k\pi >{\frac {1}{\delta }}-{\frac {\pi }{2}}\\[0.5em]&\implies 2k\pi +{\frac {\pi }{2}}>{\frac {1}{\delta }}\\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ {\frac {1}{a}}>{\frac {1}{b}}>0\iff 00\right.}\\[0.5em]&\implies \left|{\frac {1}{{\frac {\pi }{2}}+2k\pi }}\right|<\delta \\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ x={\frac {1}{{\frac {\pi }{2}}+2k\pi }}\right.}\\[0.5em]&\implies \left|x\right|<\delta \end{aligned}}}

{\displaystyle {\begin{aligned}\left|f(x)-f(0)\right|&=\left|\sin \left({\frac {1}{x}}\right)-0\right|\\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ x={\frac {1}{{\frac {\pi }{2}}+2k\pi }}\right.}\\[0.5em]&=\left|\sin \left({\frac {\pi }{2}}+2k\pi \right)\right|\\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ \sin \left({\frac {\pi }{2}}+2k\pi \right)=1\right.}\\[0.5em]&=\left|1\right|\geq {\frac {1}{2}}=\epsilon \end{aligned}}}

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

## Exercises

### Epsilon-delta criterion: Signum function

Exercise (Discontinuity of the signum function)

Prove that the signum function is ${\displaystyle \operatorname {sgn} :\mathbb {R} \to \mathbb {R} }$ is discontinuous:

${\displaystyle \operatorname {sgn}(x)={\begin{cases}1&x>0\\0&x=0\\-1&x<0\end{cases}}}$

How to get to the proof? (Discontinuity of the signum function)

In order to prove discontinuity of the entire function, we just have to find one single argument where it is discontinuous. Considering the graph of ${\displaystyle \operatorname {sgn} }$ , we can already guess, which argument this may be:

The function has a jump at ${\displaystyle x_{0}=0}$ . So we expect it to be discontinuous, there. It remains to choose an ${\displaystyle \epsilon >0}$ that makes it impossible to find a ${\displaystyle \delta >0}$ , that makes the function fit into the ${\displaystyle 2\epsilon }$-${\displaystyle 2\delta }$-rectangle. This is done by setting ${\displaystyle \epsilon }$ smaller than the jump height ${\displaystyle 1}$ - for instance ${\displaystyle \epsilon ={\tfrac {1}{2}}}$. For that ${\displaystyle \epsilon }$, no matter how ${\displaystyle \delta >0}$ is given, there will be function values above or below the ${\displaystyle 2\epsilon }$-${\displaystyle 2\delta }$-rectangle.

So let ${\displaystyle \delta >0}$ be arbitrary. We need to show that there is an ${\displaystyle x}$ with ${\displaystyle |x-x_{0}|<\delta }$ but ${\displaystyle |f(x)-f(x_{0})|\geq \epsilon }$ . Let us take a look at the inequality ${\displaystyle |x-x_{0}|<\delta }$ :

${\displaystyle |x-x_{0}|<\delta {\stackrel {x_{0}=0}{\iff }}|x|<\delta }$

This inequality classifies all ${\displaystyle x}$ that can be used for the proof. The particular ${\displaystyle x}$ we choose has to fulfill ${\displaystyle |f(x)-f(x_{0})|\geq \epsilon }$ :

{\displaystyle {\begin{aligned}{\begin{array}{rrl}&|f(x)-f(x_{0})|&\geq \epsilon \\[0.5em]\iff &|\operatorname {sgn}(x)-\operatorname {sgn}(x_{0})|&\geq {\frac {1}{2}}\\[0.5em]\iff &|\operatorname {sgn}(x)-\operatorname {sgn}(0)|&\geq {\frac {1}{2}}\\[0.5em]\iff &|\operatorname {sgn}(x)|&\geq {\frac {1}{2}}\end{array}}\end{aligned}}}

So our ${\displaystyle x}$ needs to fulfill both ${\displaystyle |x|<\delta }$ and ${\displaystyle |\operatorname {sgn}(x)|\geq {\tfrac {1}{2}}}$ . The second inequality ${\displaystyle |\operatorname {sgn}(x)|\geq {\tfrac {1}{2}}}$ may be achieved quite easily: For any ${\displaystyle x\neq 0}$ , the value ${\displaystyle \operatorname {sgn}(x)}$ is either ${\displaystyle 1}$ or ${\displaystyle -1}$. So ${\displaystyle x\neq 0}$ does always fulfill ${\displaystyle |\operatorname {sgn}(x)|=1\geq {\tfrac {1}{2}}}$.

Now we need to fulfill the first inequality ${\displaystyle |x|<\delta }$. From the second inequality, we have just concluded ${\displaystyle x\neq 0}$ . This is particularly true for all ${\displaystyle x}$ with ${\displaystyle 0 . Therefore, we choos ${\displaystyle x}$ to be somewhere between ${\displaystyle 0}$ and ${\displaystyle \delta }$ , for instance ${\displaystyle x={\tfrac {0+\delta }{2}}={\tfrac {\delta }{2}}}$.

The following figure shows that this is a sensible choice. The ${\displaystyle 2\epsilon }$-${\displaystyle 2\delta }$-rectangle with ${\displaystyle \epsilon ={\tfrac {1}{2}}}$ and ${\displaystyle \delta ={\tfrac {1}{2}}}$ is drawn here. All points above or below that rectangle are marked red. These are exactly all ${\displaystyle x}$ inside the interval ${\displaystyle (-\delta ,\delta )}$ excluding ${\displaystyle 0}$. Our chosen ${\displaystyle x={\tfrac {\delta }{2}}}$ (red dot) is situated directly in the middle of the red part of the graph above the rectangle:

So choosing ${\displaystyle x={\tfrac {\delta }{2}}}$ is enough to complete the proof:

Proof (Discontinuity of the signum function)

We set ${\displaystyle x_{0}=0}$ (this is where ${\displaystyle f}$ is discontinuous). In addition, we choose ${\displaystyle \epsilon ={\tfrac {1}{2}}}$. Let ${\displaystyle \delta >0}$ be arbitrary. For that given ${\displaystyle \delta }$, we choose ${\displaystyle x={\tfrac {\delta }{2}}}$. Now, on one hand there is:

{\displaystyle {\begin{aligned}{\begin{array}{rrl}&{\frac {1}{2}}&<1\\[0.5em]\implies &{\frac {1}{2}}\delta &<\delta \\[0.5em]\implies &\left|{\frac {1}{2}}\delta \right|&<\delta \\[0.5em]\implies &\left|{\frac {1}{2}}\delta -0\right|&<\delta \\[0.5em]\implies &|x-x_{0}|&<\delta \end{array}}\end{aligned}}}

But on the other hand:

{\displaystyle {\begin{aligned}|f(x)-f(x_{0})|&=|\operatorname {sgn}(x)-\operatorname {sgn}(x_{0})|\\[0.5em]&=\left|\operatorname {sgn} \left({\frac {\delta }{2}}\right)-\operatorname {sgn}(0)\right|\\[0.5em]&=\left|1-0\right|\\[0.5em]&=1\\[0.5em]&\geq {\frac {1}{2}}=\epsilon \end{aligned}}}

So indeed, ${\displaystyle \operatorname {sgn} }$ is discontinuous at ${\displaystyle x_{0}=0}$ . Hence, the function ${\displaystyle \operatorname {sgn} }$ is discontinuous itself.