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2 questions on continuity/continuous extensions

  • Thread starter akoska
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  • #1
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If D is a dense subset of R and f is uniformly continuous, prove that f has a continuous extension to R.

I said:

if x0 is in D, then f(x0) is continuous.

Let x0 be in R\D. D dense in R--> there exists {x_n} in D s. t. {x_n}--> x0.

{x_n} is a Cauchy sequence converging to x. f is uniformly continuous on D--> {f(x_n)} is Cauchy and thus converges to y.

Define f(x0)=y

Then I'm stuck. I want to show that f(x0) is continuous, but I'm not sure how.

Another question: Prove that if D is closed (instead of dense) , then f also has a continuous extension

I'm not sure even where to begin with this question.
 

Answers and Replies

  • #2
Dick
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For the first one, just use the definition of continuity. For the second the extension is clearly not unique so you just need to define one. Hint: the complement of a closed set in R is a countable union of open intervals.
 
  • #3
HallsofIvy
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If D is a dense subset of R and f is uniformly continuous, prove that f has a continuous extension to R.

I said:

if x0 is in D, then f(x0) is continuous.
f(x0) is a number. It makes no sense to say that a number is continuous. Perhaps you meant to say that f is continuous at x0.

Let x0 be in R\D. D dense in R--> there exists {x_n} in D s. t. {x_n}--> x0.

{x_n} is a Cauchy sequence converging to x. f is uniformly continuous on D--> {f(x_n)} is Cauchy and thus converges to y.
Do you need uniformly continuous for that?

Define f(x0)=y

Then I'm stuck. I want to show that f(x0) is continuous, but I'm not sure how.

Another question: Prove that if D is closed (instead of dense) , then f also has a continuous extension

I'm not sure even where to begin with this question.
 
  • #4
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f(x0) is a number. It makes no sense to say that a number is continuous. Perhaps you meant to say that f is continuous at x0.


Do you need uniformly continuous for that?

yes, sorry, that's what I meant.

No, you don't. I simply did not want to type an almost identical question out again, and then forgot to mention that you don't need uniform continuity.

So, using the definition of continuity, every sequence {x_n} in D that converges, {f(x_n)} converges. But this only takes into account {x_n} in D, right? So what about {x_n} in R? Is it that: Given a converging sequence {x_n} in R that converges to x0, we can find an equivalent sequence {x_n} in D that converges to x0 because x0 must be a limit point of D or a point in D (consequence of D dense)? So f is continuous on R.

I'm still unclear on what to do in the closed set quesion.
 
  • #5
Dick
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Because D is dense if you have a cauchy sequence if R, you can find a cauchy sequence in D with the same limit. In light of what I said about the second question you only have to define f in a continuous way across open intervals. If f(0)=0 and f(1)=1 how would you define f on (0,1) to make it continuous?
 

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