Showing that a limit does not exist

[Mr Davis 97]

Homework Statement

Let $f(x) = 0$ if $x$ is rational and $=1$ of $x$ is irrational. Prove that $\lim_{x\to a} f(x)$ does not exist for any $a$.

Homework Equations

The Attempt at a Solution

I need help setting this one up. I was thinking that maybe I can argue by contradiction and suppose that there exists an $a\in \mathbb{R}$ such that $\lim_{x\to a} f(x) = L$ for some $L$. I don't know what the contradiction would then be though.

EDIT: Can I assume that $L$ is either 1 or 0? If so I think I know how to proceed.

 

[fresh_42]

Homework Statement

Let $f(x) = 0$ if $x$ is rational and $=1$ of $x$ is irrational. Prove that $\lim_{x\to a} f(x)$ does not exist for any $a$.

Homework Equations

The Attempt at a Solution

I need help setting this one up. I was thinking that maybe I can argue by contradiction and suppose that there exists an $a\in \mathbb{R}$ such that $\lim_{x\to a} f(x) = L$ for some $L$. I don't know what the contradiction would then be though.

EDIT: Can I assume that $L$ is either 1 or 0? If so I think I know how to proceed.

Yes. $|f(x)-L|\in \{\,|1-L|\, , \,|L|\,\}$ which can only get arbitrary small, if $L \in \{\,0\, , \,1\,\}$ for otherwise there will be a constant gap.

 

[Mr Davis 97]

Yes. $|f(x)-L|\in \{\,|1-L|\, , \,|L|\,\}$ which can only get arbitrary small, if $L \in \{\,0\, , \,1\,\}$ for otherwise there will be a constant gap.

So would an argument go like this?

Let $a\in\mathbb{R}$.
The only two possible values for a limit of $\lim_{x\to a}f(x)$ are $0,1$, since otherwise there will always be a constant gap and hence $f$ would not get arbitrarily small.

However, $\lim_{x\to a}f(x) \not = 1$ for the following reason: Set $\epsilon = 1/2$ and let $\delta > 0$. Then there exists a rational number $x_0$ in $0<|x_0-a|<\delta$ by the density of the rationals. However, $|f(x_0)-1| = |0-1| = 1 \not < 1/2$, and so $\lim_{x\to a}f(x) \not = 1$.

Also, $\lim_{x\to a}f(x) \not = 0$ for the following reason: Set $\epsilon = 1/2$ and let $\delta > 0$. Then there exists a irrational number $x_0$ in $0<|x_0-a|<\delta$ by the density of the irrationals. However, $|f(x_0)-0| = |1-0| = 1 \not < 1/2$, and so $\lim_{x\to a}f(x) \not = 0$.

 

[scottdave]

I am not sure what it's called, but I think there is a theorem stating something like between any 2 rational numbers, there is an uncountable infinite amount of irrational numbers. The uncountable may be the key.

 

[fresh_42]

Density is a good key. For any $\delta > 0$ you can always find a rational number $r$ and an irrational number $j$ such that $|r-j| < \delta$. If the limit exists, then $x \longmapsto f(x)$ is continuous at $a$. Now prove that it is not continuous by the $\varepsilon-\delta$ definition.

 

[Mr Davis 97]

Density is a good key. For any $\delta > 0$ you can always find a rational number $r$ and an irrational number $j$ such that $|r-j| < \delta$. If the limit exists, then $x \longmapsto f(x)$ is continuous at $a$. Now prove that it is not continuous by the $\varepsilon-\delta$ definition.

So does my proof work or not? If not then why not?

 

[fresh_42]

Your wording of what I had written is a bit strange. The principle is o.k., but I think you will have to start with the distinction of $a$ being rational or not, simply because in what you have written, you cannot rule out $a=x_0$ and with it $f(x_0)=f(a)\,.$ I still find you should work with the negation of the $\varepsilon-\delta$ definition for practicing.

 

[member 587159]

Easier: use sequences of rationals/irrationals approaching irrationals/rationals together with the sequential characterisation of limits.

 

[Ray Vickson]

I am not sure what it's called, but I think there is a theorem stating something like between any 2 rational numbers, there is an uncountable infinite amount of irrational numbers. The uncountable may be the key.

What I have always found mysterious is the fact that between any two rationals there is an irrational and between any two irrationals there is a rational. Nevertheless, there are a lot more irrationals than rationals!