Is Every Continuous Open Mapping Monotonic?

[malicx]

Homework Statement

So I'm going through baby Rudin. Problem 15 in chapter 4 has us trying to prove that every continuous open mapping is monotonic. I'm trying to see how this is the case. So, I'm considering

f(x) = sin(x). Let V = (Pi/3, 7*Pi/12) be an open set. Then, f(V) = ($$\sqrt{3}$$/2, (1 + $$\sqrt{3}$$/2$$\sqrt{2}))$$. But, f is not monotonic on this interval so I must have the image must not be open (since sin(x) is certainly continuous). So, what did I miss here?

Open maps take open sets to open sets V -> f(V)

Thanks!

 

[malicx]

I was mistaken about f(V) before, here is what we have:

x takes values in R, but f(X) = [-1, 1].

So, f(V) = (sqrt[3]/2, 1]. But that is still open in [-1, 1].

so I am still missing something I think...

 

[╔(σ_σ)╝]

I don't understand what you wrote. f(V) is neither open or closed.

 

[malicx]

I don't understand what you wrote. f(V) is neither open or closed.

The complement of f(V) is [-1, sqrt(3)/2], obviously closed, so f(V) is open

 

[╔(σ_σ)╝]

The complement of f(V) is [-1, sqrt(3)/2], obviously closed, so f(V) is open

That is not the complement. The complement is [-1, sqrt(3)/2).

 

[malicx]

That is not the complement. The complement is [-1, sqrt(3)/2).

f(V) = (sqrt(3)/2, 1]

f(V)^c = {x in [-1, 1] | x not in f(V)}.

sqrt(3)/2 is not in f(V), so it IS in f(V)^c. Thus, [-1, sqrt(3)/2].

If that were the complement, then f(V) union f(V)^c would not be the entire interval... which is a basic complement law. right?

Call U the universe of discourse. then for any A in U, A union A^c = U

 

[╔(σ_σ)╝]

The way I understand closed sets is that every point in the set has a neighbourhood lying COMPLETLY in the set. Every neighbourhood of 1 has points outside the set in your example.

 

[╔(σ_σ)╝]

f(V) = (sqrt(3)/2, 1]

f(V)^c = {x in [-1, 1] | x not in f(V)}.

sqrt(3)/2 is not in f(V), so it IS in f(V)^c. Thus, [-1, sqrt(3)/2].

If that were the complement, then f(V) union f(V)^c would not be the entire interval... which is a basic complement law. right?

Call U the universe of discourse. then for any A in U, A union A^c = U

f(V) complement is = ( - infinity, sqrt(3)/2] union (1, infinity).

If we intersects f(V) with [-1,1] then it becomes closed.

 

[malicx]

The way I understand closed sets is that every point in the set has a neighbourhood lying COMPLETLY in the set. Every neighbourhood of 1 has points outside the set in your example.

1 is the endpoint of the set, there ARE no points greater than 1 in im(f).

EDIT: the DOMAIN is R, the IMAGE is [-1, 1]

 

[╔(σ_σ)╝]

1 is the endpoint of the set, there ARE no points greater than 1 in im(f).

Which means that there is a neighbourhood of 1 that contains points not in im(f).

Okay, I am tired of arguing; perhaps I am wrong. regardless, it's almost 1am and I should probably get some sleep.

Hopefully, someone else can help with your question. :-)

 

[╔(σ_σ)╝]

Btw you may want to consider the map
f(x) = 1/(x^2 +1) and how it takes an open interval (-1,1) to ( 1/2 , 1] which by MY definition is not an open interval in R.

 

[malicx]

Btw you may want to consider the map
f(x) = 1/(x^2 +1) and how it takes an open interval (-1,1) to ( 1/2 , 1] which by MY definition is not an open interval in R.

if you are considering the codomain to be (1/2, 1], then since your image is equal to the codomain it is both open AND closed by the very definition of a topology (the null set and the whole set are both open and closed)... but we aren't considering them in R, we are considering only (-1, 1) and (1/2, 1] the latter of which is somehow "different" from R. If you let the codomain be R, of course (1/2, 1] isn't open. I think you may be confused about the definition of an "open map."

Call a function f: X -> Y an open map if for any open set U in X, the image f(U) is open in Y. *Notice, in particular, that you must absolutely define what you want your codomain to be, because it WILL change whether the image f(U) is open or not!

Take a look here, I was having the same confusion as you last week, https://www.physicsforums.com/showthread.php?p=2973748#post2973748

But all of this is neither here nor there. I think I am missing something a little more subtle than what you suggest.

 

[╔(σ_σ)╝]

Yes you are correct. Sorry about the confusion.