Do these "hyper-discontinuous" functions exist?

Question

Suppose $\ X\subset \mathbb{R}\ $ and $\ Y\subset \mathbb{R}.$

Definition: A function $\ f: X \to Y\ $ is hyper-discontinuous if for every $\ x\in X,\ \ \exists\ \delta>0,\ \varepsilon>0\ $ such that $\ y\in X \setminus \{x\},\ \vert y-x \vert < \delta,\ \implies \vert f(x) - f(y) \vert \geq \varepsilon.$

Hyper-discontinuous is a term I just made up, but it seems appropriate, because $\ f\ $ is hyper-discontinuous implies $\ f\ $ is nowhere continuous, whereas the converse must be false, and nowhere continuous is the concept most closely relating to "most discontinuous function" that I am aware of.

My questions are the following:

1. Is there a hyper-discontinuous function $\ f:[0,1] \to [0,1]\ ?$
2. Is there a hyper-discontinuous function $\ f:[0,1]\cap\mathbb{Q} \to [0,1]\cap\mathbb{Q}\ ?$

I've spent a while on these questions, but keep getting confused.

I think Blumberg's theorem might be related to this, but I'm not sure.

Answer 1

There is no hyper-discontinuous function $f:[0,1]\to\mathbb{R}$

Proof: (1) there exist $\delta_0>0$ and $\epsilon_0>0$ such that for uncountably many $x\in[0,1]$, the condition in the definition is fulfilled with $\delta>\delta_0$ and $\epsilon>\epsilon_0$.

(2) there is an interval of length $\delta_0/2$ that contains uncountably many $x$ as in (1)

(3) $f$ cannot be defined on the interval from (2) without contradiction

There is a hyper-discontinuous function $f:[0,1]\cap \mathbb{Q}\to [0,1]\cap \mathbb{Q}$.

Proof: define $f(q/p)=1/p$, where $q/p$ is in lowest terms.