Analysis (left and right-hand limits, monotonicity)

[raphile]

Homework Statement

Assume that $f$ is a monotone increasing function defined on $\mathbb{R}$ and that for some $x_0\in \mathbb{R}$ the left and right limit coincide. Can you prove that $f$ is continuous at $x_0$? Either give a complete proof or a counterexample.

Homework Equations

The Attempt at a Solution

I believe the function should be continuous at $x_0\text{ }$ (otherwise, my whole answer is incorrect!). This is the proof I've come up with:

Given that the left and right limits coincide at $x_0$, call this limit $L$. Then we know (by definitions of left and right limits):

$\forall \epsilon>0 \text{ }\exists \delta_1>0: \forall x\in \mathbb{R}\text{ with } -\delta_1<x-x_0<0\text{ we have }|f(x)-L|<\epsilon$
$\forall \epsilon>0 \text{ }\exists \delta_2>0: \forall x\in \mathbb{R}\text{ with } 0<x-x_0<\delta_2\text{ we have }|f(x)-L|<\epsilon$

Suppose (for a contradiction) that the function is not continuous at $x_0$. Then we can say that $|f(x_0)-L|=\tau>0$, where $\tau$ is some positive number. If $f(x_0)>L$ then we have $f(x_0)-L=\tau$, and if $f(x_0)<L$ then we have $L-f(x_0)=\tau$.

In the case where $f(x_0)-L=\tau$, take $\epsilon=\tau\text{ }$ in the right-hand limit definition. Then we can say that for $x\in (x_0,x_0+\delta_2)$ (in particular, for $x=x_0+\delta_2/2$) we have $|f(x)-L|<\tau$ and hence $f(x_0+\delta_2/2)<L+\tau$. This means we have $f(x_0)=L+\tau$ but $f(x_0+\delta_2/2)<L+\tau$, which contradicts the statement that $f$ is monotonically increasing.

In the case where $L-f(x_0)=\tau$, take $\epsilon=\tau\text{ }$ in the left-hand limit definition. Then we can say that for $x\in (x_0-\delta_1,0)$ (in particular, for $x=x_0-\delta_1/2$) we have $|f(x)-L|<\tau$ and hence $f(x_0-\delta_1/2)>L-\tau$. This means we have $f(x_0)=L-\tau$ but $f(x_0-\delta_1/2)>L-\tau$, which contradicts the statement that $f$ is monotonically increasing.

I'd really like someone to check whether this proof makes sense or possibly if there is a simpler way to do it, or even if I have the wrong answer to start with. Thanks!

 

[Sajet]

Yes, this works. Sure, you could shorten the second part if you like. For instance you can use that by monotony f(x₀) is stuck between two converging sequences with the same limit point, i.e.

$f(x_0-1/n) \leq f(x_0) \leq f(x_0+1/n)$ for all n

$\Rightarrow \lim_{n \in \mathbb N} f(x_0-1/n) \leq f(x_0) \leq \lim_{n \in \mathbb N} f(x_0+1/n)$

$\Rightarrow L \leq f(x_0) \leq L \Rightarrow f(x_0) = L$

but yours is fine, too.

Also, depending on your definition of continuity (epsilon-delta-criterion?), you might want to point out how all of this fits together in proving the continuity of f.