Continuity of a function and bound

Question

Please, check my solution. Task: let $f(x): \mathbb{R} \rightarrow \mathbb{R}$. If $f(x)\le M$ for every $x\in\mathbb{Q}$ then is is also $\le M$ for every irrational number. Me solution: if it is continuous then for every $\varepsilon > 0$ we can choose $\delta > 0$ such that for every $x, y \in\mathbb{R}$ $|x - y| < \delta \Rightarrow |f(x) - f(y)| < \varepsilon$. Ok, it is obvious that we can choose $x\notin\mathbb{Q}$ and $y\in\mathbb{Q}$. Then if $f(a) > M$ for every irrational number then $f(x) = M + \varepsilon'$ for some $\varepsilon' > 0$. Then $|f(x) - f(y)| > \varepsilon'$, so it means that our function is not continuous. Contradiction.

Answer 1

I disagree with Kavi Rama Murthy that your argument hardly makes any sense, but it is very poorly stated and also has some technical issues. Let me go through some of the problems:

• Wall of text: Spacing and line breaks is very useful to readers. When we see someone can't be bothered to help us out by putting them in, we can't be expected to bother helping them, either. If your post had been a little longer, I would have passed it by without reading because you ignored this.

Task: let $f(x): \Bbb R \to \Bbb R$. If $f(x)\le M$ for every $x\in \Bbb Q$, then it is also $\le M$ for every irrational number.

• As stated, this is false. For example, it does not hold for the characteristic function of the irrationals: $$f(x) = \begin{cases} 0 & x \in \Bbb Q\\1 & x \notin \Bbb Q\end{cases}$$

Note that nowhere does it say that $f$ is continuous. Since you argued from continuity, your "contradiction" doesn't contradict anything in the problem. If the actual problem does not state that $f$ is continuous, then your argument is false.

In the rest, I am going to assume that $f$ being continuous was actually part of the problem statement that you left off. Don't ever do that. When you ask for help, you need to give people all the relevant information.

• You are trying to prove $$\text{if }f: \Bbb R \to \Bbb R\text{ is continuous and }\le M \text{ on }\Bbb Q\text{ then it is }\le M\text{ on all of }\Bbb R.$$ But what your logic actually proves is $$\text{if }f: \Bbb R \to \Bbb R\text{ is }\mathit{uniformly}\text{ continuous and }\le M\text{ everywhere except possibly}\\\text{ for one point }x\notin \Bbb Q\text{, then it is }\le M\text{ on all of }\Bbb R.$$

Where you went wrong:

if it is continuous then for every $\varepsilon > 0$ we can choose $\delta > 0$ such that for every $x, y \in\Bbb R$ $|x - y| < \delta \implies |f(x) - f(y)| < \varepsilon$.

This is a statement of uniform continuity. Note that $\delta$ can only depend on the value of $\epsilon$. It must hold true for all $x, y \in \Bbb R$.

The statement of normal continuity on $\Bbb R$ is $$\text{For every }x \in\Bbb R\text{ and }\varepsilon > 0\text{, there is a }\delta > 0\text{ such that for all }y\in\Bbb R, \\|x - y| < \delta \implies |f(x) - f(y)| < \varepsilon$$ Note that here $\delta$ can depend on both $\varepsilon$ and $x$, and need only work for all $y \in \Bbb R$.

The other place you went wrong is:

Ok, it is obvious that we can choose $x\notin \Bbb Q$ and $y \in \Bbb Q$. Then if $f(a) > M$ for every irrational number then $f(x) = M + \varepsilon'$ for some $\varepsilon' > 0$.

You are trying a proof by contradiction, where you assume the opposite of the conclusion, and show this requires the hypothesis to be false. But the conclusion is $$f(x) \le M\text{ for all irrational numbers }x.$$ The opposite of that is NOT $$f(x) > M\text{ for all irrational numbers }x.$$ The opposite is $$\text{there exists an irrational number }x \text{ with }f(x) > M.$$

By choosing an arbitrary irrational $x$, and assuming $f(a) > M \forall a \notin \Bbb Q$, your argument only contradicts uniform continuity and being unbounded at a single point.