Question

Let a function f :R to R be defined by
f(x)={2if x belong to Q,4 if x does not belong to Q

CHECK WHETHER f is continuous on B
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Check whether f is continuous on B.

Answer 1

Let a function ,

$f:\R\rightarrow \R$ defined by

$f(x)=\begin{cases} 2 & \text{if} & x\in Q\\ 4 &\text{if} &x\notin Q \end{cases}$

Claim: $f$ is discontinue at every point of $\R$ .

Case:1

Let $a$ be any rational number so that $f(a)=2$ .

Then the neighborhood of $(a-\frac{1}{n},a+\frac{1}{n})$ of $a$ contains an irrational number $a_n \ \text{for each} \ n\in\N$ .Since irrational number are dense in $\R$ .

i,e , $a_n\in(a-\frac{1}{n},a+\frac{1}{n}) , \ \forall \ n\in \N$

$\implies a-\frac{1}{n}<a_n<a+\frac{1}{n} \ , \ \forall \ n\in \N$

$\implies \mid a_n -a\mid<\frac{1}{n} \ , \forall \ n\in \N$ .

$\implies \mid a_n-a\mid \rightarrow0 \ \text{as} \ n\rightarrow \infin$

$\implies a_n-a \rightarrow0 \ \text{as} \ n\rightarrow\infin$

$\implies a_n\rightarrow a \ \text{as} \ \ n\rightarrow \infin$

$\implies (a_n) \ \text{converges to } \ a$ .

Now , $(a_n)=4 \ ,\forall \ n\ \text{as} \ a_n$ is irrational and $f(a)=2$ as $a$ is rational.

$\therefore \text{lim}_{n\to \infin} f(a_n)=4\neq f(a).$

i,e, $(f(a_n))$ does not converges to $f(a) \ \text{when} \ a_n\rightarrow a.$

Thus , $f$ is discontinuous at all rational points $a.$

Case:2

Let $b$ be any irrational number so that $f(b)=4.$

Since rational number are also dense in real number. As explained above ,we can choose a rational number $b_n$ such that

$\mid b_n-b\mid<\frac{1}{n} ,\ \forall \ n\in \N$ .

$\implies b_n\rightarrow b \ \text{as} \ n\rightarrow \infin$ .

$\implies (b_n) \ \text{converges to } \ b$

Now, $f(b_n)=2 ,\ \forall \ n\in\N \ \text{as} \ b_n$ is rational and $f(b)=4.$

$\therefore \text{lim}_{x\to\infin} f(b_n)=2\neq f(b).$

$\implies (f(b_n))$ does not converges to $f(b) \ \text{when} \ b_n\rightarrow b.$

Thus,the function $f$ is discontinuous at all irrational points $b$ .

Hence the given function $f$ is discontinuous at all point of $\R$ .

So in particular on any set $B$ .