# Proof about a theorem that says a function is continuous if only if f is right continuous in a and left continuous.

Theorem:

A function $f:D\to \mathbb R$ is continuous in $a\in D$ $\iff f$ is left and right continuous in $a$.

Proof:

I firstly thought just to write down the definitions of left and right continuous and then it trivially shows the theorem. But apparently it isn't sufficient.

Let $f:D\to\mathbb R$ and consider $a\in D$. The function $f$ is rightcontinuous in $a \iff$ $$(\forall\epsilon\gt 0)(\exists\delta\gt 0)(a\le x\lt a+\delta\Rightarrow |f(x)-f(a)|\lt\epsilon)$$

and left continous $\iff$ $a \iff$ $$(\forall\epsilon\gt 0)(\exists\delta\gt 0)(a -\delta\lt x\le a\Rightarrow |f(x)-f(a)|\lt\epsilon)$$

So I found a proof online on this webpage.

My question is there another way to prove this maybe with the use of my definitions? I'd most appreciate it.

• Hint: Let $\varepsilon > 0$ and choose $\delta = \min \{ \delta_1 , \delta_2 \}$. – Gonzalo Benavides Aug 24 '18 at 11:06
• And then use $\delta_1$ for RC and $\delta_2$ for LC? Ok, then. Let's say I replace them with that. How can I logically combine those too? – Anonymous196 Aug 24 '18 at 11:14
• So RC would become $(\forall\epsilon\gt 0)(\exists\delta_1\gt 0)(a\le x\lt a+\delta_1\Rightarrow |f(x)-f(a)|\lt\epsilon)$ and LC is $(\forall\epsilon\gt 0)(\exists\delta_2\gt 0)(a-\delta_2\lt x\le a\Rightarrow |f(x)-f(a)|\lt\epsilon)$. Why should you choose for the min of those $\delta$'s? – Anonymous196 Aug 24 '18 at 11:23

## 2 Answers

We show that if $f$ is left and right continuous, it is continuous. (The other direction is easier, can you do it?)

Let $f$ be left and right continuous at $a$. Let $\epsilon>0$. There are $\delta_l>0$ and $\delta_r>0$ with $a-\delta_l<x\le a\Longrightarrow |f(x)-f(a)|<\epsilon$ and $a\le x <a+\delta_r\Longrightarrow |f(x)-f(a)|<\epsilon$.

Now as the hint said, choose $\delta=\min(\delta_l,\delta_r)$. Then if $|x-a|<\delta$, either $x\le a$ or $x\ge a$. In the first case, $x>a-\delta>a-\delta_l$ and so $|f(x)-f(a)|<\epsilon$. In the second case, $x<a+\delta<a+\delta_r$ and so $|f(x)-f(a)|<\epsilon$ and the function is continuous.

The textbook solution you showed uses an equivalent characterisation of convergence using sequences.

• Well I don't know how can I do the other direction besides the stuff I've already posted. – Anonymous196 Aug 24 '18 at 21:33

@Kusma:

$\fbox {$\Rightarrow$}$:

Let $f$ be continuous in $a$. Let $\epsilon\gt 0$. Then there exists a $\delta\gt 0$, so $|x-a|\lt\delta\Longrightarrow |f(x)-f(a)|\lt\epsilon$

So to divide this into LC and RC we choose a $\delta:= min(\delta_1,\delta_2)$. Then because of $|x-a|\lt\delta$ this implies that $x\le a$ or $x\ge a$ and we have $x\gt a-\delta\gt a-\delta_1$ and so $|f(x)-f(a)|\lt\epsilon$ for LC.

The other case is $x\lt a+\delta\lt a+\delta_2$ and so $|f(x)-f(a)|\lt\epsilon$ for RC.

Which will give us the definitions for LC and RC above.

Is this correct?

• Can I assume the upvoter is you,@Kusma? – Anonymous196 Aug 25 '18 at 12:38