Continuous and Disc. functions

[Unassuming]

I am trying to understand continuous and discontinuous. These are two assignments I have for a class. I am just looking for some feedback...


Let A= {1/n : n is natural}

Then, f(x)= (x , if x in A)
(0 , if x not in A)

This is discontinuous on A but continuous on A complement?



Let B = { x : 0 \leq x < 1 }

Then, g(x) = (1-x , 0\leqx<1 and x is rational)
(0 , otherwise)

And this is discontinuous on B but continuous on B complement?

 

[Dick]

Are you just guessing? For the first one, 0 is in A complement. Is f continuous at 0? For the second one, maybe. But why do you think so? You should give reasons as well as just a guess at the answer.

 

[Unassuming]

I am trying to understand continuous and discontinuous. These are two assignments I have for a class. I am just looking for some feedback...


Let A= {1/n : n is natural}

Then, f(x)= (x , if x in A)
(0 , if x not in A)

This is discontinuous on A but continuous on A complement?



Let B = { x : 0 \leq x < 1 }

Then, g(x) = (1-x , 0\leqx<1 and x is rational)
(0 , otherwise)

And this is discontinuous on B but continuous on B complement?

I think that the problem point for set A is 0. I believe that f(x) is continuous there because I drew an epsilon in my mind around f(x) on the y-axis and imagined that I could find a delta such that not only the (x not in A) values would map into it (obviously) but that I could choose delta such that the 1/n values were far enough down, and thus in the epsilon N. I hope I am right on this. The problem points are hard to figure out.

With B I am using the same reasoning. This time though we have discontinuous at x=0 because the f(x) values are at value 1 while there are always values in your delta N such that f(x)=0. But at x=1, f(x) is getting as close as we want it to get to f(x)=0.

 

[Dick]

Now that you've explained your reasoning, I feel better better about agreeing with you. I think that's right.