Do continuous functions preserve open intervals?

[buddyholly9999]

Let $$f: D \rightarrow \mathbb{R}$$ 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

 

[berkeman]

"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

 

[buddyholly9999]

?

$$D \subseteq \mathbb{R}$$

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.

 

[berkeman]

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.

 

[buddyholly9999]

that's exactly what i meant...am i misunderstanding something?

 

[Muzza]

(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.

 

[mathwonk]

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...

 

[buddyholly9999]

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...

 

[nocturnal]

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

 

[plweber]

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.

 

[plweber]

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