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I am trying to understand the solution to the following question.

At which $c\in\mathbb{R}$ is the function $f:\mathbb{R}\rightarrow\mathbb{R}$ defined by $$f(x)=\begin{cases}x&\text{if $x$ is rational,}\\1-x&\text{if $x$ is irrational,}\end{cases}$$ continuous?

The solution states that the only answer is $c=1/2$, but am not sure why this is so.

I have thought about what might happen if we suppose (for a contradiction) that $f$ is continuous at some $c\neq1/2$. To arrive at a contradiction, I am trying to use the fact that $\exists$ an irrational number between any two numbers. However, I have not yet managed to arrive at a contradiction.

Could I have some suggestions as to how to progress, please?

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For my feeling the whole question is a bit hypothetical. It helps understanding the distribution of rational and irrational numbers and the definition of continuous functions, though.

Per definition, for a continuous function $f(x)$, you can provide a $\delta$ for any given $\varepsilon$, so that $f(x-d)-f(x)<\varepsilon$ for all $d \in [-\delta...\delta]$. More simple, you can add little random deviations to $x$ and you will only get litte random deviations in $f(x)$.

The problem with this function definition is: For any given $\delta$, even the smallest, you will find both rational and irrational numbers between $x$ and $x+\delta$. So, for any $\varepsilon$ smaller $|(x) - (1-x)|$, no $\delta$ can be found to fulfill the requirement.

The requirement for continuous functions can only be satisfied where $x$ and $1-x$ converge. As both are linear functions, they converge only once.

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  • $\begingroup$ @ philipp Why have you written 'for any $\epsilon$ smaller $|(x)-(1-x)|$? I would have thought that this should read 'for any $\epsilon$ smaller than $|(x)-(1-c)|\ldots$', where $c\in[-\delta, \delta]$. Please could you explain? $\endgroup$ – Caleb Owusu-Yianoma Apr 15 '15 at 13:22
  • $\begingroup$ $x$ and $1-x$ are the two definitions of $f(x)$, so $|f(x_1)-f(x_2)|$, where $x_1$ is rational and $x_2$ irrational, but both are virtually the same, is $|x-(1-x)|$ $\endgroup$ – philipp Apr 15 '15 at 13:49
  • $\begingroup$ @ philipp By 'both are virtually the same', do you mean that $x_1$ and $x_2$ are almost equal to each other? $\endgroup$ – Caleb Owusu-Yianoma Apr 15 '15 at 14:22
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This is because at $x = \frac{1}{2}$ we have $f(x) = \frac{1}{2} = 1 - \frac{1}{2} = 1 - x$

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  • $\begingroup$ Alright. However, how do we know whether or not $f$ is continuous at some other point $c\neq1/2$? $\endgroup$ – Caleb Owusu-Yianoma Apr 10 '15 at 18:13
  • $\begingroup$ Think of it this way: We know that the real numbers are dense, with the rationals being countably infinite on any interval, and the irrationals being uncountably infinite. At $\frac{1}{2}$, there are an infinitely dense amount of $x$ with corresponding images $1-x$ in the neighborhood of $f(\frac{1}{2})$. When we aren't at $\frac{1}{2}$, we have an infinite dense interval of points mapping onto $1-x$, and a countable set of points mapping onto $x$, therefore not continuos. $\endgroup$ – Addison Apr 10 '15 at 19:26
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    $\begingroup$ @CKKOY Suppose that your function is continuous at some $c$. Then: $$\begin{align} \lim_{\substack{r\rightarrow c \\ r\in \Bbb Q}} f(r) & = \lim_{\substack{x\rightarrow c \\ x\in \Bbb R -\Bbb Q}} f(x) &= f(c)\\ \lim_{\substack{r\rightarrow c \\ r\in \Bbb Q}} r &= \lim_{\substack{x\rightarrow c \\ x\in \Bbb R -\Bbb Q}} 1-x & \\ c &= 1-c \end{align}$$ So $c=1/2$ $\endgroup$ – Kitegi Apr 10 '15 at 19:27
  • $\begingroup$ @Farnight This was a very helpful response. $\endgroup$ – Caleb Owusu-Yianoma Apr 11 '15 at 18:18

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