Is Function f Continuous Only at Zero?

[miren324]

Here's the problem:
Let f(x)={x, x in Q; 0, x in R\Q.
Show f is continuous at c if and only if c = 0.
Hint: You may want to use the following theorem: Let A and B be two disjoint subsets of R and f₁:A[tex]\rightarrow[/tex]R and f₂:B[tex]\rightarrow[/tex]R. Define f:A[tex]\cup[/tex]B[tex]\rightarrow[/tex]R by

f(x)={f₁(x), x in A; f₂(x), x in B​

Let c be an accumulation point of both A and B and L in R. Show c is an accumulation point of A[tex]\cup[/tex]B and L = limx[tex]\rightarrow[/tex]cf(x) if and only if L = limx[tex]\rightarrow[/tex]cf₁(x) = limx[tex]\rightarrow[/tex]cf₂(x).

Here's what I tried. Let f(x)=x for x in Q be f₁ and f(x)=0 for x in R\Q be f₂. Then by definition of continuous functions I have for all epsilon>0 there exists delta>0 such that |f(x)-f(c)|<epsilon when x is in Q[tex]\cup[/tex]R\Q and |x-c|<delta.

This leads to two possibilities: x is in Q or x is in R\Q, so we have |f(x)-f(c)|=|x-f(c)|<epsilon and |f(x)-f(c)|=|0-f(c)|<epsilon.

The problem is, what if c is not 0, but rather is x in Q? Then |x-f(c)|=|x-x|<epsilon which is true. Then I have f is continuous at c with c being nonzero. However, I know that I am supposed to PROVE this, not DISPROVE it. Am I missing something here?

 

[HallsofIvy]

I have no idea what you mean by "what if c is not 0, but rather is x in Q". c is a fixed number, a constant. x is a variable.

c is rational or it is irrational.

1) If c is rational, then f(c)= c. If c is not 0, take [itex]\epsilon= |c|/2> 0[/itex]. For any [itex]\delta> 0[/itex], there exist an irrational x such that [itex]|x- c|< \delta[/itex]. Since x is irrational, [itex]f(x)= 0[/itex]. Then [itex]|f(x)- f(0)|= |0- c|= |c|> |c|/2= \epsilon[/itex].

2) If c is irrational, then f(c)= 0. If c is not 0, take [itex]\epsilon= |c|/2> 0[/itex]. For any [itex]\delta> 0[/itex], there exis a rational x such that [itex]|x- c|< \delta[/itex]. Since x is rational, [itex]f(x)= c[/itex]. Then [itex]|f(x)- f(0)= |c- 0|= c> |c|/2= \epsilon[/itex].