A counter-example?

**ModusPonens** · May 15th 2012, 07:05 AM

Hello

I'm working on the exercises of Munkres' Topology. Section 30, exercise 1-a says: "Show that in a first-countable $T_1$ space, every one-point set is a $G_{ \delta}$ set"

Now, I've seen the solution to this exercise and I believe it's wrong, because it assumes that a countable basis at x is a family of sets that get "arbitrarily small", that is, it's like a countable collection of balls of radius 1/n (I know the space doesn't need to be metrizable, I'm just giving the intuition). This assumption is wrong because there could be a strange topology in $R$ in which the neighbourhoods of 0 would be intervals of the form (-1-1/n,1+1/n). This is a countable basis at 0.

Now the counter example: $R$ with the finite complement topology. In this topology, it is $T_1$ and first countable, but no one-point set is a $G_{ \delta}$ set.

Is this counter-example right? Thanks in advance

**Plato** · May 15th 2012, 07:24 AM

Originally Posted by ModusPonens

Munkres' Topology. Section 30, exercise 1-a says: "Show that in a first-countable $T_1$ space, every one-point set is a $G_{ \delta}$ set"

Now, I've seen the solution to this exercise and I believe it's wrong, because it assumes that a countable basis at x is a family of sets that get "arbitrarily small", that is, it's like a countable collection of balls of radius 1/n (I know the space doesn't need to be metrizable, I'm just giving the intuition). This assumption is wrong because there could be a strange topology in $R$ in which the neighbourhoods of 0 would be intervals of the form (-1-1/n,1+1/n). This is a countable basis at 0.

Now the counter example: $R$ with the finite complement topology. In this topology, it is $T_1$ and first countable, but no one-point set is a $G_{ \delta}$ set.

"An example of a space which is not first-countable is the cofinite topology on an uncountable set (such as the real line)."

**ModusPonens** · May 15th 2012, 08:36 AM

Thank you, but I don't understand why. Why isn't the family $\{ B_n : B_0=R- \{ x+1 \} ;B_n=B_{n-1}- \{x+n \} \}$ a countable basis at x?

And if you can explain that, please explain how to prove the original question.

**Plato** · May 15th 2012, 08:57 AM

Originally Posted by ModusPonens

Thank you, but I don't understand why. Why isn't the family $\{ B_n : B_0=R- \{ x+1 \} ;B_n=B_{n-1}- \{x+n \} \}$ a countable basis at x?

How do you define a local basis?
And why do you think the above is one?

**ModusPonens** · May 15th 2012, 09:22 AM

I will transcribe the definition from the book.

A space X is said to have a countable basis at x if there is a countable collection B of neighborhoods of x such that each neighborhood of x contains at least one of the elements of B. A space that has a countable basis at each of its points is said to satisfy the first countability axiom, or to be first-countable.

I think that the family I defined is a countable family of neighborhoods of x, such that each neighborhood contains at least one other (in fact infinite neighborhoods). Since x is arbitrary, that would make the space first-countable.

PS: Neighborhood is defined just as an open set containing x, and not a set which contains an open set containing x.

**Plato** · May 15th 2012, 10:29 AM

I will simply prove that $\mathbb{R}$ with the cofinite topology cannot be first countable. Suppose that $a\in\mathbb{R}$ and $\{O_n\}$ is a local basis at $a$
$\forall n,~\mathbb{R}\setminus O_n$ is finite.
Thus $[\bigcup\limits_n {\mathbb{R}\backslash {O_n}}$ must be countable.
So $\bigcap\limits_n {{O_n}}$ must be uncountable and contain a point $b$ such $b\ne a$ .
But $Q=\mathbb{R}\setminus\{b\}$ is an open set and $a\in Q$ .
Now it is impossible for $O_n\subset Q$ .
Contradiction .

**Plato** · May 15th 2012, 11:29 AM

Originally Posted by ModusPonens

Munkres' Topology. Section 30, exercise 1-a says: "Show that in a first-countable $T_1$ space, every one-point set is a $G_{ \delta}$ set"

Now, I've seen the solution to this exercise and I believe it's wrong, because it assumes that a countable basis at x

Of course it it assumes that a countable basis at x because that is what first countable means.

**ModusPonens** · May 15th 2012, 02:16 PM

I'm not sure if you read my first post without attention or if I wasn't clear, but my contention is that a countable basis at x is not a collection of sets that gets arbitrarily small. That's why it's difficult to do the exercise.

Help would be apreciated. No time constraints, this isn't a homework assignment.

**ModusPonens** · May 15th 2012, 02:43 PM

Originally Posted by Plato

I will simply prove that $\mathbb{R}$ with the cofinite topology cannot be first countable. Suppose that $a\in\mathbb{R}$ and $\{O_n\}$ is a local basis at $a$
$\forall n,~\mathbb{R}\setminus O_n$ is finite.
Thus $[\bigcup\limits_n {\mathbb{R}\backslash {O_n}}$ must be countable.
So $\bigcap\limits_n {{O_n}}$ must be uncountable and contain a point $b$ such $b\ne a$ .
But $Q=\mathbb{R}\setminus\{b\}$ is an open set and $a\in Q$ .
Now it is impossible for $O_n\subset Q$ .
Contradiction .

Is it a contradiction because $\{O_n\}\bigcup\{Q\}$ should form a countable basis and it doesn't?

If that is so, why does there have to be another countable basis at $a$ . One for each point is enough to make it first-countable.

I'm sorry if I'm bothering you, but I can't get this out of my mind.

**ModusPonens** · May 19th 2012, 05:14 PM

How about this as a counterexample to the original question: $\mathbb{R}$ with the following subbasis: $\{\mathbb{R}-\{x\}:x\in\mathbb{R}\}\cup\{(k-1/n,k+1+1/n):k\in\mathbb{Z},n\in\mathbb{N},n>10\}$

**ModusPonens** · May 19th 2012, 05:22 PM

In this topology, the set $\{1/2\}$ is not a $G_{\delta}$ set.

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