Proof involving topological spaces and density.

[Slats18]

Homework Statement

Let (A,S) and (B,T) be topological spaces and let f : A -> B be a continuous function. Suppose that D is dense in A, and that (B,T) is a Hausdorff space. Show that if f is constant on D, then f is constant on A.

Homework Equations

D is a dense subset of (A,S) iff the intersection of D and U is not the empty set, for every non-empty open set U.
D is dense on A, i.e., the closure of D is A.
(B,T) is Hausdorff, i.e., for any two points in B, there exist disjoint open sets U,V in T such that the intersection of U and V is the empty set.

The Attempt at a Solution

Suppose, by way of contradiction, that this is not true. Then, f is not constant on A, if f is constant on D.

I'm not sure how I go about this, I can't assume f(D) is dense in B because we haven't been told it's onto, only continuous, correct? How can I show f is constant on D?

 

[Dick]

Ok, so let c be the constant such that f(D)=c. If f is not constant on A, then there is an element a such that f(a)=d where d is not equal to c. Ok, so far? Now what does f being continuous at a tell you? Now add 'Hausdorff'.

 

[Slats18]

The function f is said to be continuous at a iff for each open set V containing f(a), there is an open set U containing a such that f(U) is a subset of V.

So there is some subset U that contains a, and by the denseness of D in A, there must be some element of D in the subset U. But because f is not constant for a, but is for D, this is a contradiction as the subset U cannot be disjoint of itself...?

I got some help today showing that if you take a point in A, apply the function to get two points in B (as f is only continuous), show that in Hausdorff spacwe that there is open subsets, and by continuity these subsets are in A, and then that because of denseness, these is a point from D in both of these, but both can't go to the one constant point. Roughly the same idea?

 

[Dick]

That's not terribly clear. But since B is Hausdorff there is an open subset V of d that does not contain c. So there is an open subset U of a such that f(U) is contained in V, yes? So f(U) does not contain c. Now what does density tell you?

 

[Slats18]

Density tells us that, if D is dense in A, there is an element of D in every open subset of A.

So, there is an element d in U. Then, f(U) is the function on the subset U, which contains an element of D, so f(U) contains c. QEA

Sound about right?

 

[Dick]

Density tells us that, if D is dense in A, there is an element of D in every open subset of A.

So, there is an element d in U. Then, f(U) is the function on the subset U, which contains an element of D, so f(U) contains c. QEA

Sound about right?

There's fragments of the proof in there. Can you state it clearly in the form of an argument by contradiction?

 

[Slats18]

Well, I'll try and remember what I handed in, it's gone now so it's not too much of a worry, now I'm focused on Cauchy's Integral Formula haha.

Skip question...

Suppose, by way of contradiction, that this is not true. Then, f is not constant on A, if f is constant on D.

If f is not constant on A, then there are two points x,y$$\in$$A, x$$\neq$$y, such that f(x)=c and f(y)=d, with c,d$$\in$$B as A$$\rightarrow$$B.

Then, by (B,T) being Hausdorff, there exists non-empty, disjoint open sets W,Z$$\in$$B, W$$\cap$$Z= the null set, such that c$$\in$$W and d$$\in$$Z.

As the function is continuous, for each open set W,Z$$\in$$B containing points c and d respectively, there is an open set U,V$$\in$$A containing x and y respectively, such that f(U)$$\subseteq$$W and f(V)$$\subseteq$$Z.

Now, as D is dense in A, then D$$\cap$$U$$\neq$$ the null set and D$$\cap$$V$$\neq$$ the null set, i.e., there is an element of D in every open set of A. We denote these points in D$$\cap$$U and D$$\cap$$V as m and n, respectively.

Finally, by the function f being constant on D, f(m)=f(n), which infers that f(m),f(n)$$\in$$W, or f(m),f(n)$$\in$$Z, but not both, which contradicts Hausdorff space.


That's basically the gist of it, it was a lot rougher in the one I handed in though haha.

 

[Dick]

Well, I'll try and remember what I handed in, it's gone now so it's not too much of a worry, now I'm focused on Cauchy's Integral Formula haha.

Skip question...

Suppose, by way of contradiction, that this is not true. Then, f is not constant on A, if f is constant on D.

If f is not constant on A, then there are two points x,y$$\in$$A, x$$\neq$$y, such that f(x)=c and f(y)=d, with c,d$$\in$$B as A$$\rightarrow$$B.

Then, by (B,T) being Hausdorff, there exists non-empty, disjoint open sets W,Z$$\in$$B, W$$\cap$$Z= the null set, such that c$$\in$$W and d$$\in$$Z.

As the function is continuous, for each open set W,Z$$\in$$B containing points c and d respectively, there is an open set U,V$$\in$$A containing x and y respectively, such that f(U)$$\subseteq$$W and f(V)$$\subseteq$$Z.

Now, as D is dense in A, then D$$\cap$$U$$\neq$$ the null set and D$$\cap$$V$$\neq$$ the null set, i.e., there is an element of D in every open set of A. We denote these points in D$$\cap$$U and D$$\cap$$V as m and n, respectively.

Finally, by the function f being constant on D, f(m)=f(n), which infers that f(m),f(n)$$\in$$W, or f(m),f(n)$$\in$$Z, but not both, which contradicts Hausdorff space.


That's basically the gist of it, it was a lot rougher in the one I handed in though haha.

I don't think that really hits the point very accurately. The only real point you need to make out of all of that is that c is not contained in f(V). Why? But V contains points of D. Hence c is contained in f(V). Why? Now that's a contradiction of f not being constant.