Answer to Question #109841 in Real Analysis for Pappu Kumar Gupta

Question #109841
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
Ï
Î
=
Q
Q
4, if x
2, if x
f x
Check whether f is continuous on B.
1
Expert's answer
2020-04-15T17:50:32-0400

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" .


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS