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))$$\cap$$f(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 definition 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).