Show that $f$ is continuous at $\frac12$

Question

Let $f$: $[0,1] \to [0,1]$ be defined by

$$f(t):=\begin{cases} t& \text{if}\; x \in \Bbb{Q}\\1-t &\text{otherwise}\end{cases}$$ I have to show that $f$ is only continuous at $t=$$\frac12$

I know the Def: $f$ is said to be continuou at $x$ if for every $\epsilon>0$, exists a $\delta$ such that $|f(x)-f(y)|<\epsilon$ for every $y$ belonging to $(x-\delta, x+\delta)$

and Def: $f$ is said to be continuous at $x$ if $\lim_{n \to\infty}f(a_n)=f(x)$ for every $\lim_{n \to\infty}a_n=x$

How can I show this?

Answer 1

Verify that for every $t\in[0,1]$ you have $$ \frac12 - \left|t-\frac12\right| \leq f(t) \leq \frac12 + \left|t-\frac12\right| . $$ The LHS and RHS are both continuous at $\frac12$ and have the same limit.

Alternatively, from the given inequality you deduce also $$ \left|f(t)-\frac12\right| \leq \left|t-\frac12\right|, $$ which implies that $\lim_{t\to1/2} f(t)=1/2$.

Answer 2

Whatever $x \in [0,1]$ you take, we have $$\vert f(t)-f(1/2)\vert=\vert t-1/2\vert$$

So take $\delta =\varepsilon$ to conclude the desired result!

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Suppose $f$ is continuous at other points, say $f$ is continuous at $x \in \Bbb{Q} \cap [0,1]\setminus\{1/2\}$

Take $x_n \in \Bbb{Q^c}$ so that $x_n \rightarrow x $. By continuity $f(x_n) \rightarrow f(x)=x$. But $f(x_n)=1-x_n \rightarrow 1-x$. But $1-x \neq x$. so we have a contradiction! similarly check this for irrational points in $[0,1]$