Continuity of Finite Set f: R → R - Proofs

[C.E]

1. Let X be R be a finite set and define f : R [tex]\rightarrow[/tex] R by f(x) = 1 if x [tex]\in[/tex] X and f(x) = 0 otherwise. At which points c in R is f continuous? Give proofs.

3. I don't know how to start this, do you think it is ok to assume that X represents an interval of R? If not how can you possibly deduce the points continuity?

 

[quasar987]

X is finite. Meaning it contains a finite number of points. So X is certainly not an interval.

Try some concrete examples with increasing degree of complexity. Say X={0}. What then? (i.e., where is f continuous?) Now what if X={-1,1}, etc. If you've solved the problem in these two particular cases, then surely you can guess the answer to the general case and back your intuition with a proof.

 

[C.E]

Is this right?

The function is discontinous for all x in X and continuous elsewhere.
To prove discontinuity at x in X let x_1, x_2, ... x_n be the points in X then if we assume X_2 is the member of X closest to x_1. Then taking episilon =0.5 and only considering delta less than 0.5|x_1 - x_2| we prove discontinuity.

To prove continuity elsewhere we use a similar argument letting X_1 be the closest member of X to the point x not in X then setting delta= 0.5|X_1-x| completes the proof.

Any comments?

 

[quasar987]

You seem to have set x=x_1 in the first paragraph but never said so explicitly, which is confusing. Also, when you say "Then taking epsilon =0.5 and only considering delta less than 0.5|x_1 - x_2| we prove discontinuity.", I suspect that you have the right idea, but your sentence expresses it poorly. How about instead: "Then, taking epsilon=0.5, notice that for delta=0.5|x_1 - x_2|, we have 0<|x_1-y|<delta implies |f(x_1)-f(y)|=|1-0|=1>epsilon, thus proving that f is discontinuous at x_1."

In the second paragraph, I suggest adding "then for any epsilon>0, take delta= 0.5|X_1-x|, thus completing the proof.", but it can't hurt to write things more explicitely either.

 

[HallsofIvy]

I think simpler is: since X is finite, there exist [itex]\epsilon> 0[/itex] such that the distance between any two points in X is greater than [itex]\epsilon[/itex].