Continuity and countable density

[radou]

Continuity and "countable density"

Homework Statement

Let f : X --> Y be a continuous function. If X has a countable dense subset A, then f(X) has a countable dense subset, too.

The Attempt at a Solution

Since A is countable dense in X, Cl(A) = X. Since f is continuous, f(Cl(A)) = f(X) is contained in Cl(f(A)). I have a hunch that f(A) is the countable dense subset of f(X).

Assume y is in Cl(f(A)). Then every neighborhood V of y intersects f(A). I somehow need to show that y is in f(X), since then f(X) = Cl(f(A)), and hence f(A) is countable dense in f(X).

But here's where I get a bit stuck, any suggestions? If y is in f(A), it is obviously in f(X). Assume y is not in f(A). I'm trying to show that for every such y there's some x in X such that y = f(x).

 

[ystael]

"Countable density" is not a property; a countable dense subset is a subset which is both countable and dense.

Since A is countable dense in X, Cl(A) = X. Since f is continuous, f(Cl(A)) = f(X) is contained in Cl(f(A)). I have a hunch that f(A) is the countable dense subset of f(X).

OK. Stop right here. At this point you need to prove that f(A) is (1) countable and (2) dense in f(X).

(1) should be easy to see.

Before you go making any detailed argument for (2), ask yourself: What is the definition of "f(A) is dense in f(X)"? Does it look familiar?

 

[micromass]

Sadly enough, it will not be true (in general) that f(X)=Cl(f(A)). But, it is true that

f(X)=Cl(f(A))\cap f(X)

obivously. Now, does theorem 17.4 page 95 give us anything interesting?

 

[radou]

ystael, thanks for the quick reply.

"Countable density" is not a property; a countable dense subset is a subset which is both countable and dense.

I'm perfectly aware of that, I was only being a bit informal. :)

OK. Stop right here. At this point you need to prove that f(A) is (1) countable and (2) dense in f(X).

(1) should be easy to see.

Before you go making any detailed argument for (2), ask yourself: What is the definition of "f(A) is dense in f(X)"? Does it look familiar?

OK, as I stated, countability is trivial.

For (2), f(A) is dense in f(X) if Cl(f(A)) = f(X). I could assume it isn't to arrive at a contradiction, that was my aim. So, assume there is some y in Cl(f(A)) which doesn't equal f(x) for any x in X.

 

[radou]

Sadly enough, it will not be true (in general) that f(X)=Cl(f(A)). But, it is true that

f(X)=Cl(f(A))\cap f(X)

obivously. Now, does theorem 17.4 page 95 give us anything interesting?

Oh, it gives us immediately that the closure of f(A) in f(X) equals Cl(f(A))\capf(X)!

 

[ystael]

Edit: The below was a reply to your second-last comment, so I see that you have already seen the problem.

Nope. Stop. You're thinking too hard. :)

You already observed that f(X) = f(\overline{A}) \subset \overline{f(A)}.

Now, just now, you said "f(A) is dense in f(X) if \overline{f(A)} = f(X)." This definition is actually slightly incorrect. To see how it is incorrect, let Y = \mathbb{R}, X = \mathbb{R} \setminus \{0\}, A = \mathbb{Q} \setminus \{0\}, f(x) = x be the inclusion map X \to Y. Here f(A) is certainly dense in f(X), but \overline{f(A)} \neq f(X).

Based on this example, correct your definition, and you should see that you are already done.

 

[radou]

OK, I'll think about this a bit later, since I have to go. Btw, the only definiton I was using was the one from Munkres, i.e.

Definition. A subset A of a space X is said to be dense in X is Cl(A) = X.

 

[radou]

I've done a bit research, and it seems that the definition of "density" I gave applies only to metric spaces, right?

In a general topological space X, a subset A of X is dense in X if any x in X either belongs to A or is a limit point of A.

A limit point of a set A is a point x such that every of its neighborhoods intersect A in a point other than x itself.

Now, in our case, let y be a point in f(X). Suppose y doesn't belong to f(A). Let's show it is a limit point of A. Let V be a neighborhood of y in f(X). Since f is continuous, its inverse image U is open in X, and contains no points of A. Take any point x in U. Since A is countably dense in X, x is a limit point of A, so U intersects A at some point other than x. But then f(U) = V intersects f(A) in some point other than y itself, and hence y is a limit point of f(A). So, f(A) is dense in f(X). (Countably dense, by what we have shown earlier)

 

[ystael]

No -- the definition "A \subset X is dense in X if \overline{A} = X" is correct for general topological spaces. The catch is that you always need to be perfectly clear what the ambient space is in which you take the closure.

Also -- again, the phrase "countably dense" is not used, because the quality of countability is a property of the set, not of its density property.

 

[radou]

No -- the definition "A \subset X is dense in X if \overline{A} = X" is correct for general topological spaces. The catch is that you always need to be perfectly clear what the ambient space is in which you take the closure.

Also -- again, the phrase "countably dense" is not used, because the quality of countability is a property of the set, not of its density property.

I was using "A is countably dense in X" as a synonym for A is a countable subset of X, which is dense in X. Just to shorten things up. In this situation I thought there would be no disambiguity, so I am being a bit inprecise.

 

[radou]

Btw, now I realize that my conclusion from post #8 was false, since Cl(A) = A U A', where A is the set of all the limit points of A, and if A is a subset of X, and if it is dense in X, then every x in X is either in A or a limit point of A, hence X = Cl(A).