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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).

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    $\begingroup$ Well you would have to start by defining some probability measure on this set of functions. $\endgroup$ – air Sep 9 '15 at 19:01
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    $\begingroup$ depending on your concept of "most", you get different answers... $\endgroup$ – user251257 Sep 9 '15 at 19:07
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    $\begingroup$ Which probability measure are you working with? The phrase like with probability 1 is meaningless until we assign a probability law, and vague notion of randomness often results in paradoxical dead end like in the case of Bertrand paradox. $\endgroup$ – Sangchul Lee Sep 9 '15 at 19:07
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    $\begingroup$ There is another notion of "most". One can show that the set of functions which are differentiable somewhere is meager (Baire first category). $\endgroup$ – zhoraster Sep 9 '15 at 19:27
  • $\begingroup$ When I was thining about the problem I was also confused about how one would randomly choose a function in this case. Thanks for the link to bertrand's paradox, it explained the ambiguity very well. $\endgroup$ – MCT Sep 9 '15 at 21:11
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Like the commenters have pointed out, this question is somewhat ill-defined unless you pose a probability measure on the family of functions from $[0, 1]$ to $\mathbb{R}$. However, your discussion of cardinality suggests that maybe you instead want to show that the cardinality of functions in $\mathcal{F}$ that are continuous somewhere is strictly smaller than the cardinality of $\mathcal{F}$. If this were true, then it would indeed be the case that 'most' functions in $\mathcal{F}$ are discontinuous everywhere (just like since the cardinality of the reals is greater than the cardinality of the rationals, 'most' reals are irrational).

Unfortunately this is not the case. To see this, call $\mathcal{C}$ the subset of $\mathcal{F}$ of functions that are continuous at some point. Since $\mathcal{C} \subset \mathcal{F}$, the cardinality of $\mathcal{F}$ is at least that of $\mathcal{C}$. We will show an injective mapping from $\mathcal{F}$ into $\mathcal{C}$, thus showing the cardinality of $\mathcal{C}$ is at least that of $\mathcal{F}$, and hence they have equal cardinalities.

To do this, consider the map $\pi: \mathcal{F} \rightarrow \mathcal{C}$ that sends $f \in \mathcal{F}$ to the function $g$ defined by

$$g(x) = \begin{cases}f(2x) \mbox{ if } 0 \leq x \leq 1/2 \\ x \mbox{ if } 1/2 < x\end{cases}$$

Note that $g$ is continuous at $1$ (and many more values), so it belongs to $\mathcal{C}$. Furthermore this map is injective. It follows that $|\mathcal{C}| \geq |\mathcal{F}|$, and hence $|\mathcal{C}| = |\mathcal{F}|$.

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  • $\begingroup$ Do these sets both have cardinality $\aleph_1$? $\endgroup$ – MCT Sep 9 '15 at 21:07
  • $\begingroup$ @Soke they are both $2^{\aleph_1}$ $\endgroup$ – user251257 Sep 10 '15 at 1:42
  • $\begingroup$ @user251257: No, they are both $2^{2^{\aleph_0}}$. $\endgroup$ – Eric Wofsey Sep 10 '15 at 8:09
  • $\begingroup$ @EricWofsey on yeah. my bad. $\endgroup$ – user251257 Sep 10 '15 at 8:27
  • $\begingroup$ @user251257, pretty sure those two are the same..? $\endgroup$ – Morgan Rogers Aug 6 '16 at 21:20
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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.

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