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"Show that the function $f: \mathbb{R} \rightarrow \mathbb{R}$, $$f(x)=\lim_{m \rightarrow + \infty}{ \lim_{n \rightarrow +\infty}{(\cos{(m! \pi x)})^n}}$$ is discontinuous at each $x \in \mathbb{R}$."

Could you give me some hint how to show this??

Ittay Weiss
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Mary Star
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    See also: http://math.stackexchange.com/questions/265199/dirichlet-function-pointwise-convergence, http://math.stackexchange.com/questions/264889/how-is-this-called-rationals-and-irrationals or http://math.stackexchange.com/questions/275974/how-to-prove-that-lim-undersetk-rightarrow-infty-lim-cosn-pi-x2k – Martin Sleziak May 05 '14 at 10:06

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Hint: Take a look at what happens if $x$ is rational (then $m!\cdot x$ is an even integer for large enough values of $m$) and what happens if $x$ is irrational.

Comments:

  1. if $x$ is rational, then it equals $\frac pq$ for some integers $p$ and $q$. This means that if $m>q$, then $x\cdot m! = \frac{1\cdot 2\cdots q \cdot (q+1)\cdots m}{q} p$. Can you see that this is an even integer?

  2. If $x$ is irrational, $m!\cdot x$ is not an integer and hence cannot be even. The value of $\cos$ of something which is not an integer multiple of $\pi$ is some value in $(-1,1)$ (it cannot be $1$). What happens to such values as you take higher and higher powers of them?

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5xum
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  • If $x$ is rational, $m! \cdot x$ is an even integer, because $m!$ is a multiple of $2$?? Since $x$ is rational, $m! \cdot x$ is an even integer, $\cos{(m! \pi x)}=1$, right? If $x$ is irrational, isn't $m! \cdot x$ also even, but not an integer? But how can I find the value of $cos$ in this case?? Is it maybe as followed? Since it is not an integer, the value of $cos$ is not equal to $1$. – Mary Star May 05 '14 at 10:02