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Thread: A counter-example?

  1. #1
    Member ModusPonens's Avatar
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    A counter-example?

    Hello

    I'm working on the exercises of Munkres' Topology. Section 30, exercise 1-a says: "Show that in a first-countable $\displaystyle T_1$ space, every one-point set is a $\displaystyle G_{ \delta}$ 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 $\displaystyle R$ 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: $\displaystyle R$ with the finite complement topology. In this topology, it is $\displaystyle T_1$ and first countable, but no one-point set is a $\displaystyle G_{ \delta}$ set.

    Is this counter-example right? Thanks in advance
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    Re: A counter-example?

    Quote Originally Posted by ModusPonens View Post
    Munkres' Topology. Section 30, exercise 1-a says: "Show that in a first-countable $\displaystyle T_1$ space, every one-point set is a $\displaystyle G_{ \delta}$ 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 $\displaystyle R$ 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: $\displaystyle R$ with the finite complement topology. In this topology, it is $\displaystyle T_1$ and first countable, but no one-point set is a $\displaystyle G_{ \delta}$ set.
    "An example of a space which is not first-countable is the cofinite topology on an uncountable set (such as the real line)."
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    Re: A counter-example?

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

    And if you can explain that, please explain how to prove the original question.
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    Re: A counter-example?

    Quote Originally Posted by ModusPonens View Post
    Thank you, but I don't understand why. Why isn't the family $\displaystyle \{ B_n : B_0=R- \{ x+1 \} ;B_n=B_{n-1}- \{x+n \} \}$ a countable basis at x?
    How do you define a local basis?
    And why do you think the above is one?
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    Re: A counter-example?

    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.
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    Re: A counter-example?

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

    Quote Originally Posted by ModusPonens View Post
    Munkres' Topology. Section 30, exercise 1-a says: "Show that in a first-countable $\displaystyle T_1$ space, every one-point set is a $\displaystyle G_{ \delta}$ 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.
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    Re: A counter-example?

    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.
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    Re: A counter-example?

    Quote Originally Posted by Plato View Post
    I will simply prove that $\displaystyle \mathbb{R} $ with the cofinite topology cannot be first countable. Suppose that $\displaystyle a\in\mathbb{R} $ and $\displaystyle \{O_n\} $ is a local basis at $\displaystyle a $
    $\displaystyle \forall n,~\mathbb{R}\setminus O_n $ is finite.
    Thus $\displaystyle [\bigcup\limits_n {\mathbb{R}\backslash {O_n}} $ must be countable.
    So $\displaystyle \bigcap\limits_n {{O_n}} $ must be uncountable and contain a point $\displaystyle b $ such $\displaystyle b\ne a $.
    But $\displaystyle Q=\mathbb{R}\setminus\{b\} $ is an open set and $\displaystyle a\in Q $.
    Now it is impossible for $\displaystyle O_n\subset Q $.
    Contradiction .
    Is it a contradiction because $\displaystyle \{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 $\displaystyle a$. 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.
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  10. #10
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    Re: A counter-example?

    How about this as a counterexample to the original question: $\displaystyle \mathbb{R}$ with the following subbasis: $\displaystyle \{\mathbb{R}-\{x\}:x\in\mathbb{R}\}\cup\{(k-1/n,k+1+1/n):k\in\mathbb{Z},n\in\mathbb{N},n>10\}$
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  11. #11
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    Re: A counter-example?

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