Here's the question:
Is the following true or false?
There is a function $f: \mathbb R \to \mathbb R$ that satisfies the following condition:
For every $a \in \mathbb R $ and $ \epsilon \gt 0 $ there is $\delta \gt 0$ such that $\left| f(x)-f(a) \right| \lt \epsilon \implies \left| x-a \right| \lt \delta $.
My initial response:
I said that this is true for any constant function, e.g. $f(x) = 0$. In this case, $\left| f(x)-f(a) \right| = \left|0-0 \right|=0 \lt \epsilon \text{ and} \left| x-a \right|=\left|0-0 \right|=0 \lt \delta $. I know that this doesn't work because in the $\left| x-a \right|$ case, $x$ may not be $0$, obviously. However, this was the last question on a test, so I kind of just guessed because I was running out of time. Now that I've had time to think about it, though, I can't seem to figure it out (i.e., whether or not it's true or false). Any and all help here is appreciated, as always. Thanks.
For every $a∈\mathbb{R}$ and $\epsilon > 0$ there is $\delta>0$ such that
$∣f(x)−f(a)∣<\epsilon \Rightarrow |x−a|<\delta$ ,
which is equivalent to $|x−a| \geq \delta \Rightarrow ∣f(x)−f(a)∣ \geq \epsilon$,
in this case, any unbounded above and below strictly increasing function on $\mathbb{R}$ is enough.
Strictly increasing shows if there are points $x_0, x_1\in\mathbb{R}$ s.t. $f(x_0) \leq f(a) - \epsilon$ and $f(a) + \epsilon \leq f(x_1)$,
then taking $\delta = \max\{|x_0-a|,|x_1-a|\}$, $|x−a| \geq \delta \Rightarrow ∣f(x)−f(a)∣ \geq \epsilon$ is satisfied.
unboundedness above and below shows there is certainly at least such $x_0$ and $x_1$ in $\mathbb{R}$. Therefore that's enough.