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Math Help - Verification of a topology

  1. #1
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    Verification of a topology

    I have to show that

    \mathcal{T} = \left\{ (a, b] : a<b, a,b \in \mathbb{R}\right\}

    is a topology. I know that this is not a topology, because it forms the basis for a topology on \mathbb{R}. But why is this the case? I am thinking of the case, for instance, (1,3] \cup (5,7]. I don't think this is in \mathcal{T}, or am I incorrect?

    Thanks in advance,

    HTale.
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    Quote Originally Posted by HTale View Post
    I have to show that

    \mathcal{T} = \left\{ (a, b] : a<b, a,b \in \mathbb{R}\right\}

    is a topology. I know that this is not a topology, because it forms the basis for a topology on \mathbb{R}. But why is this the case? I am thinking of the case, for instance, (1,3] \cup (5,7]. I don't think this is in \mathcal{T}, or am I incorrect?

    Thanks in advance,

    HTale.
    You are right. Also,
    Since \left\{ (0, n]: n \in \mathbb{N}\right\} \in \mathcal{T} , but (0, \infty) \notin \mathcal{T} , \mathcal{T} is not closed under arbitrary union.
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  3. #3
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    Quote Originally Posted by HTale View Post
    I have to show that

    \mathcal{T} = \left\{ (a, b] : a<b, a,b \in \mathbb{R}\right\}

    is a topology. I know that this is not a topology, because it forms the basis for a topology on \mathbb{R}. But why is this the case? I am thinking of the case, for instance, (1,3] \cup (5,7]. I don't think this is in \mathcal{T}, or am I incorrect?

    Thanks in advance,

    HTale.
    As chabmgph said, you're correct.

    For your information, \mathcal{T}=\{(-\infty,a) ~:~ a \in \mathbb{R}\} does constitute a topology :P
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    Quote Originally Posted by Moo View Post
    As chabmgph said, you're correct.

    For your information, \mathcal{T}=\{(-\infty,a) ~:~ a \in \mathbb{R}\} does constitute a topology :P
    Does it? How about finite intersections of elements of \mathcal{T}?
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    Quote Originally Posted by Moo View Post
    As chabmgph said, you're correct.

    For your information, \mathcal{T}=\{(-\infty,a) ~:~ a \in \mathbb{R}\} does constitute a topology :P
    Quote Originally Posted by HTale View Post
    Does it? How about finite intersections of elements of \mathcal{T}?
    If O_1, ..., O_n \in \mathcal{T}=\{(-\infty,a) ~:~ a \in \mathbb{R}\}, then O_1 = (-\infty, a_1) \text{ and } O_2 = (-\infty, a_2), ... \text{ with } a_1, a_2, ... \in \mathbb{R}
    Then \bigcap_{i=1}^{n} O_i = (-\infty, m) where m=\min\left\{ a_1, a_2, ...\right\}.
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  6. #6
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    Quote Originally Posted by chabmgph View Post
    If O_1, ..., O_n \in \mathcal{T}=\{(-\infty,a) ~:~ a \in \mathbb{R}\}, then O_1 = (-\infty, a_1) \text{ and } O_2 = (-\infty, a_2), ... \text{ with } a_1, a_2, ... \in \mathbb{R}
    Then \bigcap_{i=1}^{n} O_i = (-\infty, m) where m=\min\left\{ a_1, a_2, ...\right\}.
    OK, and also we have the analagous statement for unions, where we take the maximum. But, how about the empty set? You will never be able to get the empty set on this topology. The whole of the real line, yes but \emptyset \notin \mathcal{T}

    Also, can I form the whole of the real line if I changed that to (-\infty, a]?
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  7. #7
    Moo
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    Oops sorry ! It's \mathcal{T}=\{\emptyset~,~\mathbb{R} ~,~ (-\infty~,~a) ~:~ a \in \mathbb{R}\}
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  8. #8
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    Quote Originally Posted by HTale View Post
    I have to show that

    \mathcal{T} = \left\{ (a, b] : a<b, a,b \in \mathbb{R}\right\}

    is a topology. I know that this is not a topology, because it forms the basis for a topology on \mathbb{R}. But why is this the case? I am thinking of the case, for instance, (1,3] \cup (5,7]. I don't think this is in \mathcal{T}, or am I incorrect?
    Coming back to the original question, a set \mathcal B is said to be a base (or basis) for a topology if the collection of (arbitrary) unions of members of \mathcal B is a topology. The necessary and sufficient condition for a collection \mathcal B of subsets of a space to form a base for a topology is that the union of all the members of \mathcal B should be the whole space and that the intersection of any finite number of sets in \mathcal B should also be in \mathcal B.

    In the case of the set \mathcal T above, it's easy enough to see that any finite intersection of half-open intervals is again a half-open interval. So \mathcal T is the base for a topology (the half-open interval topology).
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  9. #9
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    Re: Topology

    Check out Chapter 2 of Part V of Topology and the Language of Mathematics (available for free here: Bobo Strategy - Topology). It gives some examples which you might find useful.
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