Let a function ,
"f:\\R\\rightarrow \\R" defined by
"f(x)=\\begin{cases}\n2 & \\text{if} & x\\in Q\\\\\n4 &\\text{if} &x\\notin Q\n\\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" .
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