# Limit in closure, topology stuff

1. Dec 23, 2007

### jostpuur

Let $A\subset X$ be a subset of some topological space. If $x\in\overline{A}\backslash A$, does there exist a sequence $x_n\in A$ so that $x_n\to x$?

In fact I already believe, that such sequence does not exist in general, but I'm just making sure. Is there any standard counter examples? I haven't seen any.

If the point x has a countable local basis, that means a set $\{U_n\in\mathcal{T}\;|\;n\in\mathbb{N}\}$ so that $x\in U_n$ for all n, that for all environments U of x, there exists some $U_n\subset U$, then the sequence $x_n$ is simple to construct.

I guess that the existence of the countable local basis is a nontrivial matter, and since the sequence comes so naturally with it, it seems natural to assume that the sequence would not exist without the countable local basis. But what kind of topological spaces don't have countable local basis for each point? And what would be an example of a point in closure, that cannot be approached by some sequence?

Last edited: Dec 23, 2007
2. Dec 23, 2007

### Hurkyl

Staff Emeritus
Let your underlying set of X consist of all countable ordinal numbers, as well as the first uncountable ordinal number. Let the topology on X be the order topology: a basis is the set of open intervals.

Let S be the subspace of countable ordinals.

Warmup exercise: Let T be the subspace of finite ordinals. Prove that the first infinite ordinal is in the closure of T.

Exercise 1: prove that any countable collection of ordinal numbers has a supremum.
Exercise 2: prove that any countable subset of S has a supremum in S.
Exercise 3: prove that the closure of S is X.
Exercise 4: conclude that this is a counterexample.

The usual set of hyperreals makes another good counterexample; I believe in that set with its order topology, countably infinite sequences converge if and only if they are eventually constant.

3. Dec 23, 2007

### morphism

Take $\bar{S}_\Omega = [0, \omega_1]$, the one-point compactification of the ordinal space of the section by the first uncountable ordinal $\omega_1$. Then $\overline{[0, \omega_1)} = \bar{S}_\Omega$ but $\omega_1 \in \bar{S}_\Omega \backslash [0, \omega_1)$ is not the limit of any sequence out of $[0, \omega_1)$.

Edit:
Too late. :tongue2:

Last edited: Dec 23, 2007
4. Dec 23, 2007

### jostpuur

What are the open intervals in X?

5. Dec 23, 2007

### morphism

(a,b) = {x : a<x<b}, where < is the ordinal ordering.

For instance (1,2) is empty, (1,3)={2}, (1, w) = {2,3,...} (w is the first infinite ordinal), and so on.

6. Dec 23, 2007

### Hurkyl

Staff Emeritus
And the rays:

L_a = { x : x < a }
R_a = { x : x > a }

I suppose the rays are enough to define a basis, since each interval with two endpoints is the intersection of two rays.

7. Dec 23, 2007

### gel

Alternatively, let X be the set of functions R->R with the topology of pointwise convergence. Let A be those functions that are 0 except on a countable subset. Then A is sequentially closed, but its closure is all of X.

8. Dec 23, 2007

### jostpuur

I believe I understood the ordinal example.

I have difficulty seeing what precisely the open sets would be in this topology. I mean I understand what's the desired property of the topology: $f_n\to f$ in the topology $\Leftrightarrow$ $f_n\to f$ pointwisely, but I don't see how such topology exists.

Last edited: Dec 23, 2007
9. Dec 23, 2007

### gel

Wikipedia has a page on the topology http://en.wikipedia.org/wiki/Pointwise_convergence" [Broken], although they don't explicitly describe the open sets.

I'm not sure what the simplest description of the open sets is, but a basis for the topology can be defined as follows. Given any $$f:\Re\rightarrow\Re$$, finite $$S\subset\Re$$ and any $$\epsilon>0$$ let $$U(f,S,\epsilon)=\left\{g\in X:|g(x)-f(x)|<\epsilon\ \forall x\in S\right\}$$. Such sets form a basis for the topology.

For any fixed f, these sets form a basis for the neighbourhoods of f. There is no countable basis because there are uncountably many finite subsets $$S\subset\Re$$.

Last edited by a moderator: May 3, 2017
10. Dec 23, 2007

### Hurkyl

Staff Emeritus
Isn't that the topology for uniform convergence? Your wiki link says that the topology for pointwise convergence is the product topology -- each element of R gets its own epsilon.

11. Dec 23, 2007

### gel

No. It would be a basis for uniform convergence if I hadn't restricted S to be finite (putting S=R gives a basis for uniform convergence).
Letting each element of S have its own epsilon will also work. It will be a bigger basis than the one I mentioned but it gives the same topology.

btw, an alternative basis is given by the sets of the form $$\left\{h\in X: f(x)<h(x)<g(x)\ \forall x\in S\right\}$$ for all finite $$S\subset\Re$$ and $$f,g\in X$$.

12. Dec 23, 2007

### jostpuur

I see. In the product topology the subbasis would be given by sets

$$U(f,x,\epsilon) = \{g\in X\;:\; |f(x)-g(x)|<\epsilon\},$$

and the basis consists of finite intersections of these.

(edit 1)

But this was probably a useless remark...

The use of the same epsilon in the intersections doesn't seem to affect the final topology, however.

(edit 2)

Or no! Does it?! I'll not say anything to it now.

(edit 3)

Some quick thinking gave me this:
Even if the same epsilon is used in the intersections, when defining the basis, we can in the end write precisely the same open sets by using different functions f. If

$$U = \bigcap_{i\in I} U(f, x_i, \epsilon_i)$$

then there exists some functions, sets S and epsilons so that

$$U = \bigcup_{k\in K} U(f_k, S_k, \epsilon_k)$$

where each set S had only one epsilon attached to it. (The sets $I$ and $S_k$ are finite.)

Last edited: Dec 23, 2007
13. Dec 23, 2007

### gel

Yes, that's exactly it.

14. Dec 23, 2007

### gel

The set $$U(f,x,\epsilon)$$ can be written as a union of such sets with any smaller epsilon, so you can always take the minimum of the finite set of different epsilons and use that in the intersection instead.