2

My friend and I were trying to prove that a function can't be discontinuous on the irrationals. His proof was this:

Let $f:\mathbb{R}\to\mathbb{R}$ be continuous on all of $\mathbb{Q}$. Then for any $\{x_n\}\subset\mathbb{Q}$ with $x_n\to r\in\mathbb{Q}$, we have $\lim_{n\to\infty}f(x_n)=f(r)$. Now choose $\{x_n\}$ such that $\{x_n\}$ is Cauchy. Then for any $\epsilon>0$ there exists an $N_0\in\mathbb{N}$ such that for all $n,\ m\ge N_0$, $|x_m-x_n|<\epsilon$.

Now in the interval $(x_n,\ x_m)$ there's an irrational $\gamma$ between them. Therefore, $\gamma$ also satisfies the definition of continuity, and so $f(x)$ is continuous for at least one $\gamma\in\mathbb{Q}^c$ and so it can't be continuous on only the rationals.

Something about this doesn't sit well with me, it feels so off, but I can't tell why. If it were this simple all along, what's with all the $F_\sigma$ and $G_\delta$ sets and Baire Spaces and Theorems and all this other complicated stuff?

Community
  • 1
jergen von mergen schmergen
  • 41

2 Answers2

1

Now in the interval $(x_n,\ x_m)$ there's an irrational $\gamma$ between them.

That's true, but there is not necessarily any single irrational number $\gamma$ that is in the interval $(x_n, x_m)$ for every $n$ and $m$. Indeed, if $\{x_n\}$ is an increasing sequence then there is no such irrational number.

So this technique will not show convergence at any irrational number.

If you must prove convergence in such a basic way, choose an arbitrary irrational number $\gamma$ and prove continuity at that particular $\gamma$.

The point of "all the $F_\sigma$ and $G_\delta$ sets and Baire Spaces and Theorems and all this other complicated stuff," I believe, is the same as the purpose of producing a whole bunch of definitions and theorems in any branch of mathematics, namely, to make it possible to prove many facts (including many that are far more difficult to prove) with less effort than in takes to get a result like this the "simple" way.

David K
  • 98,388
0

Set of all points of continuity of a continuous function is an $F_\sigma$ set and by Baire Category Theorem $\Bbb Q^c$ is not an $F_\sigma$ set

Learnmore
  • 31,062