Vtyjkyuk

Theorem:

A function $f:Dto mathbb R$ is continuous in $ain D$ $iff f$ 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 $f:Dtomathbb R$ and consider $ain D$. The function $f$ is rightcontinuous in $a iff$ $$(forallepsilongt 0)(existsdeltagt 0)(ale xlt a+deltaRightarrow |f(x)-f(a)|ltepsilon)$$

and left continous $iff$
$a iff$ $$(forallepsilongt 0)(existsdeltagt 0)(a -deltalt xle aRightarrow |f(x)-f(a)|ltepsilon)$$

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.

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 $a-delta_l<xle aLongrightarrow |f(x)-f(a)|<epsilon$ and $ale x <a+delta_rLongrightarrow |f(x)-f(a)|<epsilon$.

Now as the hint said, choose $delta=min(delta_l,delta_r)$. Then if $|x-a|<delta$, either $xle a$ or $xge a$. 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.

@Kusma:

$fbox $Rightarrow$$:

Let $f$ be continuous in $a$. Let $epsilongt 0$. Then there exists a $deltagt 0$, so $|x-a|ltdeltaLongrightarrow |f(x)-f(a)|ltepsilon$

So to divide this into LC and RC we choose a $delta:= min(delta_1,delta_2)$. Then because of $|x-a|ltdelta$ this implies that $xle a$ or $xge a$ and we have $xgt a-deltagt a-delta_1$ and so $|f(x)-f(a)|ltepsilon$ for LC.

The other case is $xlt a+deltalt a+delta_2$ and so $|f(x)-f(a)|ltepsilon$ for RC.

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

Is this correct?