How about the space of all real-valued functions which are not continuous? You can actually show that there are some input-output pairs of real numbers which satisfy the definition of continuity. In other words, you can show continuous functions exist without ever writing one down!
Proof
$(1)$ Let $F(\mathbb{R},\mathbb{R})$ denote the set of all possible single variable, real valued functions.
$(2)$ Now let $S\in F(\mathbb{R},\mathbb{R}) $ denote the set of all functions where if the difference between output values is less than $\epsilon$, then we can find a corresponding $\delta$ which will always be greater than the difference between two arbitrary input values. This is possible, we can just make up a rule where the difference between input values $x_1$ and $x_2$ is twice the difference between $f(x_1),f(x_2)$. Play around with some numbers to get the intuition.
$(3)$ Let $f\in S$. Consider two points of the function $f(x_{1})$ and $f(x_{2})$. Remember since $f\in S$, the difference $|f(x_{1})-f(x_{2})|<\epsilon$ and $|x_1-x_2|<2\delta$ by definition. This fits the definition of continuity, and so continuous functions do exist. You can modify to show its possible to have differentiable functions, monotonic etc.
What is the cardinality of $F(\mathbb{R},\mathbb{R})$? What about the set of monotone functions? What about the set of functions which are differentiable, but the derivative is not differentiable anywhere?
To conclude, you could reasonably conjecture (based on cardinality of power sets) that there are probably a $\aleph_{2}$ number of possible subsets of $F(\mathbb{R},\mathbb{R})$ we could ask ourselves what the cardinality of each of those sets are. Consider a set $X$ where each function is bounded by some constant: $|f(x)|<M$. We could choose any $M\in (1,\infty)$ to be the bound, and that means there are $\aleph_1$ such functions in $X$ alone.
