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How can I educate the continuity of $f(x)$, as $f(x)=0\ \forall x \in \mathbb{Q}$ and $f(x)=x\ \forall x \notin \mathbb{Q}$? Because there is real number between every rational number.

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Hi! I for one, am having trouble understanding what you want to ask. Firstly "educate" must not be the verb you meant to use. Secondly, you have to be more clear about the continuity you're asking about. This function is certainly not continuous on the real numbers, but it's continuous as a function restricted to the rationals, and continuous when restricted to the irrationals. –  rschwieb Oct 11 '12 at 13:56
    
sorry for my bad english but I think the question is clair. –  pourjour Oct 11 '12 at 18:32
    
The main problem is "How can I educate the continuity". The verb "educate" makes this sentence gibberish in English. From the context, it looks like you mean "disprove". –  rschwieb Oct 11 '12 at 18:35
    
I didn't knowed if the function is continuous or not, that's why I avoid using "prove" or "disprove" –  pourjour Oct 11 '12 at 18:53
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"prove or disprove" is actually a LOT better than "educate". "Educate" is just nonsense here. –  rschwieb Oct 11 '12 at 19:00

2 Answers 2

up vote 2 down vote accepted

Let us proceed by the $\epsilon$-$\delta$ definition of continuity.

It is to be shown that $$\forall \epsilon >0 \exists \delta >0: |x|<\delta \implies |f(x)-f(0)|<\epsilon$$

Since $f(0) = 0$, this comes down to $|f(x)|<\epsilon$. Now take $\delta = \epsilon$.

Suppose that $x \in \Bbb R$ has $|x| < \delta$.

Then if $x \in \Bbb Q, |f(x)| = 0 < \epsilon$. If $x \notin \Bbb Q, |f(x)| = |x| < \delta = \epsilon$.

Thence $f$ is continuous at $0$. At other points a similar case distinction can be used to show that $f$ is discontinuous.

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Here's a non-$\epsilon-\delta$ answer to complement the other answer.

Lemma: If $f$ is a continuous function on a topological space $X$, and $g$ is another continuous function such that $f(x)=g(x)$ for all $x$ in a dense subset $D$ of $X$, then $f=g$ on all of $X$.

First notice that $g(x)=x$ and $h(x)=0$ are both continuous functions on $\mathbb{R}$. Then notice that $\mathbb{Q}$ and $\mathbb{R}\setminus\mathbb{Q}$ are both dense in $\mathbb{R}$.

Now if $f$ were continuous, it would have to be equal to both $g$ and $h$, but $g$ and $h$ are obviously not equal... contradiction!

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The lemma is not hard to establish, no matter what definition of continuity you want to use. –  rschwieb Oct 11 '12 at 14:03

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