Proving discontinuity by negation of continuity definition

Question

$f(x) = \lfloor x\rfloor$
A function $f$ is continuous at $x=a$ if:
$$\forall \epsilon >0 \ \ \exists \delta>0 \ \ \ \forall x\in (a-\delta,a+\delta) \ \ \ |f(x) - f(a)|<\epsilon.$$

The negation of this is
A function $f$ is discontinuous at x=a if:
$$\exists \epsilon >0 \ \ \ \forall \delta> 0 \ \ \ \exists x\in (a-\delta,a+\delta) \ \ \ |f(x) - f(a)| \geq \epsilon.$$

Prove that $f$ is discontinuous at $x=3$.

I just get lost in what to do first. Do I consider the $|f(x) - f(a)| \geq \epsilon$ part first? Do I pick $\epsilon$ first?

This is what I tried:
Pick $\epsilon$ to be $3$.Let $\delta>0$ be arbitrary and pick $x = 1-\delta < 3+\delta$.
Then $f(1-\delta) \leq 0 $ so $f(1-\delta) - 3 \leq -3$ so $|f(1-\delta) - 3| \geq 3 = \epsilon$.
I'm not sure if I just lucked out, but was my process correct? (Pick epsilon first, pick $x$...)

Answer 1

It's in an order. One must first pick the $\epsilon$, and then for all $\delta$, pick an $x \in (a-\delta,a+\delta)$ possibly depending on $\epsilon$ and $\delta$ such that $|f(x) - f(a)| < \epsilon$.

So, while doing this question, you should have an idea of what your $\epsilon$ is, first. What does this $\epsilon$ represent, is our question. How do we decide what value it is? To do this, we should understand how $\lfloor x \rfloor$ behaves in a neighbourhood of the point $3$.

Indeed, just to the left of $3$, it takes the value $2$, and to the right it takes the value $3$. So, in a neighbourhood of three, you can find a point whose function value is always separated from three by exactly one, simply by taking a point just to the left of three. This means, that you can always find a points, whose function value is separated from three , by more than, say one-half( any fraction smaller than one would have done here). This is in words. Now let's say it in symbols.

Let $\epsilon = \frac 12$. Let $\delta > 0$ be any positive number. Then, in the interval $(3-\delta,3+\delta)$, choose the number $x = 3-\frac{\delta}{2}$. This lies just enough to the left of three, and is also contained in the interval $(3-\delta,3+\delta)$. Of course, $\lfloor x\rfloor \leq 2$ by definition, but then $\lfloor x \rfloor - \lfloor 3 \rfloor \geq 1 > \epsilon = \frac 12$. Hence, we have that $\lfloor \cdot \rfloor$ is discontinuous at $x = 3$.

You can use this to prove that $\lfloor \cdot \rfloor$ is actually discontinuous at all integers, and continuous everywhere else.

Answer 2

$\epsilon = 3$ will not work. Specifically, if you pick $\epsilon = 3$, then I pick $\delta = 0.5$, and you only have $x\in (2.5, 3.5)$ to work with. You cannot make $|\lfloor x\rfloor - \lfloor 3 \rfloor|>3$ with that. Thus for that $\epsilon$ there exists a $\delta$, which means that your $\epsilon$ is not a witness to discontinuity.

If you look at the graph, it has jumps of length $1$, so you need to pick an $\epsilon<1$, say $\frac12$.

Once you've done that, you show that no matter which $\delta>0$ I throw at you, you can find aan $x\in (3-\delta, 3+\delta)$ such that $|f(x) - f(3)|<\frac12$. For instance, you might choose $3-\frac\delta2$, for this yields $$ \left|\left\lfloor3-\frac\delta2\right\rfloor - \lfloor3\rfloor\right|\geq1>\frac12 $$ and we are done.