TITLE: How to show that "most" functions $f : [0,1] \to \Bbb R$ are discontinuous everywhere?
QUESTION [6 upvotes]: Consider the family of functions $\mathcal F$ consisting of all functions $f : [0,1] \to \Bbb R$. I think that most of these functions will be discontinuous everywhere, that is, if you were to pick a random function $f \in \mathcal F$ it will be everywhere discontinuous with probability $1$.
How can one prove this statement? My first thought was to somehow prove that the set of $f$ that are continuous at any point $x \in [0,1]$ is countable, but that isn't true (e.g. the family of constant functions in $[0,1]$ is uncountable). Perhaps show that this set is $\aleph_1$ and the original set is $\aleph_2$? (I don't know if $|\mathcal F| = \aleph_2$ is true, it's just a thought).

REPLY [3 votes]: As mentioned in the comments, this question is ill-defined unless you describe more precisely what you mean by "random" and "probability".  Here is one reasonable such definition (for a slightly different version of the question) for which it is true that a random function is discontinuous everywhere with probability $1$.
Consider only functions $f:[0,1]\to[0,1]$, and choose them randomly by independently choosing the value of $f(t)$ for each $t\in [0,1]$ with respect to the uniform distribution on $[0,1]$.  More precisely, our measure space is the Cartesian product of uncountably many copies of $[0,1]$ with Lebesgue measure, one copy for each $t\in [0,1]$.
Let $f$ be such a "randomly chosen" function; I claim that with probability $1$, $f$ is nowhere continuous.  For any interval $(a,b)\subset[0,1]$ with rational endpoints, then with probability $1$ there is some $x\in(a,b)$ such that $f(x)<1/3$.  Indeed, there are infinitely many $x\in (a,b)$, and for each one of them there is an independent $1/3$ chance that $f(x)<1/3$, so with probability $1$ there will be some $x$ such that $f(x)<1/3$ (if you flip a 3-sided coin infinitely many times, you will almost surely get at least one heads; you can prove this rigorously by a simple calculation).  Similarly, with probability $1$ there is some $y\in(a,b)$ such that $f(y)>2/3$.  There are only countably many such intervals $(a,b)$, so it follows that with probability $1$, for every interval $(a,b)$ with rational endpoints., there exist $x,y\in (a,b)$ such that $f(x)<1/3$ and $f(y)>2/3$.
But no such function can be continuous anywhere.  For if $f$ were continuous at $t\in (0,1)$, then we would be able to find some small interval $(a,b)$ with rational endpoints containing $t$ such that $|f(z)-f(t)|<1/6$ for all $z\in (a,b)$.  Choosing $x,y\in (a,b)$ such that $f(x)<1/3$ and $f(y)>2/3$, we find by the triangle inequality that $|f(x)-f(y)|\leq |f(x)-f(t)|+|f(t)-f(y)|<1/3$, which is a contradiction.