Do continuous functions preserve open intervals?

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Discussion Overview

The discussion revolves around the properties of continuous functions and their effects on sets, particularly focusing on whether continuous functions preserve open intervals. Participants explore various examples and counterexamples related to closed and open sets, compactness, and the behavior of functions defined on different types of domains.

Discussion Character

  • Debate/contested
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • Some participants propose that if D is closed, then f(D) is closed, questioning whether there are simpler counterexamples than the one provided (D={2n pi + 1/n: n in N}, f(x)=sin(x)).
  • One participant challenges the claim that D = (0, 1) is not closed, suggesting that D = N (natural numbers) with f(x) = 5 might be a better counterexample.
  • Another participant clarifies that (0, 1) is an open set, as it does not contain its bounds, and the closure of (0, 1) is [0, 1].
  • There is a discussion about the implications of the intermediate value theorem, with one participant asserting that it guarantees that a continuous function cannot yield a counterexample if D is an interval.
  • Some participants suggest using functions like x --> 1/[1+x^2] or the exponential function applied to the real line to explore the behavior of continuous functions on closed sets.
  • One participant corrects an earlier statement, indicating that if D is an interval that is not open, then f(D) can still be an interval that is not open, providing a new example with D = (0, 1] and f(x) = (sin(1/x))/x.
  • Another participant claims that for D = [1, infinity), f(D) = (0, infinity) is open, which contradicts the earlier claim about preserving the openness of intervals.

Areas of Agreement / Disagreement

Participants express differing views on the properties of continuous functions and their effects on sets. There is no consensus on the examples provided, and multiple competing views remain regarding the preservation of open intervals and the nature of closed and open sets.

Contextual Notes

Some claims depend on the definitions of closed and open sets, and the discussion includes unresolved mathematical steps regarding the behavior of functions on various domains.

buddyholly9999
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Let [tex]f: D \rightarrow \mathbb{R}[/tex] be continuous.

Is there an easier function that counterexamples;
if D is closed, then f(D) is closed
than D={2n pi + 1/n: n in N}, f(x)=sin(x) ?


Plus, these counterexamples are all the same with the domain changed, just correct me if I'm wrong.

If D is not closed, then f(D) is not closed.
CE: D = (0, 1) and f(x) = 5
If D is not compact, then f(D) is not compact.
CE: We use same CE as above
If D is infinite, then f(D) is infinite.
CE: D = all real numbers and f(x) = 5
If D is an interval, then f(D) is an interval
CE: Use same CE as first
 
Last edited:
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"If D is not closed, then f(D) is not closed.
CE: D = (0, 1) and f(x) = 5"

Why is D = (0,1) not closed? I think you mean more like
CE: D = N, and f(x) = 5
 
?

[tex]D \subseteq \mathbb{R}[/tex]

an open set is one that doesn't contain it's bounds. The closure of (0, 1) is [0, 1]...umm...i'm not sure what to say...(0, 1) is an open set.
 
Fair enough, I guess I misunderstood the notation. I thought you were saying that the domain of the function was all the numbers between 0 and 1.
 
that's exactly what i meant...am i misunderstanding something?
 
(0, 1) is indeed (extremely) standard notation for the set {x; 0 < x < 1}, which is not closed.

Is there an easier function that counterexamples;
if D is closed, then f(D) is closed
than D={2n pi + 1/n: n in N}, f(x)=sin(x) ?

What about D = the natural numbers and f(x) = 1/x?

If D is an interval, then f(D) is an interval
CE: Use same CE as first

But f(D) = {5} = [5, 5]. The intermediate value theorem garantuees that you can't find a counterexample.
 
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the continuous image of a compacts et is always compact, so you must use a closed set that is not bounded. say x-->1/[1+x^2] applied to the (closed) real line.

or the exponential function applied to the real line. or arctan applied to the real line.

etc...
 
Muzza said:
But f(D) = {5} = [5, 5]. The intermediate value theorem garantuees that you can't find a counterexample.

Oh yeah...haha..I typed that one up wrong. It was supposed to be "if D is an interval that is not open, then f(D) is an interval that is not open"

my bad...
 
buddyholly9999 said:
Oh yeah...haha..I typed that one up wrong. It was supposed to be "if D is an interval that is not open, then f(D) is an interval that is not open"

my bad...
how about D = (0, 1] and f(x) = (sin(1/x))/x
 
  • #10
nocturnal said:
how about D = (0, 1] and f(x) = (sin(1/x))/x

That function would have f(D)=(-infinity,infinity) or, in other words, the reals. This interval is both open and closed so it is not a counterexample.
 
  • #11
I got it!

This is for: if D is an interval that is not open, then f(D) is an interval that is not open.

Take f:D->Reals

f(x)=(x*sin(x))+x+(1/x)

D=[1,infinity) which is not open

f(D)=(0,infinity) which is open

Woot! Woot!

Patrick
South Dakota State University - Real Analysis I
 
Last edited:

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