Is the Pre-Image Always Open in Topological Spaces?

[Jooolz]

Hi all,

I am struggling with the following:

If X and Y are topological spaces. and f: X x Y → ℝ is a continuous function (product topology on X x Y, Euclidean topology on ℝ)

Let g: X → ℝ defined by g(x) = sup { f(x,y) | y in Y }

Then: If A=(r, ∞) for r in ℝ, g^-1(A) is open. And If A=(-∞, t) for t in ℝ, g^-1(A) is not always open.

Why is that? How can I know if g^-1(A) is open or not if I don't know anything about X??

Does anyone have an idea?

kind regards,

 

[zhentil]

This is one of those "apply the definitions" problems. Let x \in g^{-1}(t,\infty). By definition, there exists \epsilon >0 such that g(x)>t+2\epsilon. By definition of supremum, there exists y \in Y such that f(x,y)>t+\epsilon. By definition of continuity, there exists an open set A containing (x,y) such that f(x,y)>t+\epsilon/2 for all (x,y) \in A. By definition of the product topology, there exists an open set B in X containing x such that \forall z \in B, \exists y_z \in Y, (z,y_z) \in A. Hence g(z) > t for all z in B.

For the reverse, just construct a counterexample. Take the plane and consider the function f(x,y)=arctan(x^2 y^2)

 

[Jooolz]

Hi zhentil,

Thank you very much for helping!

Sorry, I don't see immediately why this holds:

"By definition of continuity, there exists an open set A containing (x,y) such that
f(x,y)>t+ϵ/2 for all (x,y)∈A"

For this we need that some subset of { f(x,y) } is open?

 

[Jooolz]

or is it the set were t + ε/2 < f(x,y) < t + ε ?

 

[zhentil]

(t+\epsilon/2,\infty) is an open set in R. Hence its pre-image is open by definition of continuity. (x,y) lies in that set.

 

[Jooolz]

thank you.

"For the reverse, just construct a counterexample. Take the plane and consider the function
f(x,y)=arctan(x²y²)"

Would g^-1((-∞, b)) be open if Y is a compact space?

 

[Bacle2]

Hi all,

I am struggling with the following:

If X and Y are topological spaces. and f: X x Y → ℝ is a continuous function (product topology on X x Y, Euclidean topology on ℝ)

Let g: X → ℝ defined by g(x) = sup { f(x,y) | y in Y }

Then: If A=(r, ∞) for r in ℝ, g^-1(A) is open. And If A=(-∞, t) for t in ℝ, g^-1(A) is not always open.

Why is that? How can I know if g^-1(A) is open or not if I don't know anything about X??

Does anyone have an idea?

kind regards,

Well, let's see. Let me review for next time I teach point set topology, so you can
combine it with Zhentil's answer:

What is g^-1(a,b)? it is the collection of all x such that there is a y in Y
with a<f(x,y)<b.

We have g=Sof(x,y) , where f:XxY→f(XxY), and S:f(XxY)→ℝ , and g^-1:=

f^-1os^-1.

Like Zhentil said, if there is y₀ with f(x,y₀)>a , then , by

(assumed) continuity of f, there is a ball B(x,y₀) where f(x,y)>a . This

gives you openness in the subspace f(XxY). Now, compose with f^-1, to

get an open set in XxY, by assumed continuity of f(x,y).