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Math Help - Simple Topology Proof

  1. #1
    Super Member Aryth's Avatar
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    Simple Topology Proof

    Suppose U and V are pointsets of \mathbb{R}. If U and V both have a leftmost point, prove that the union U \cup V also has a leftmost point.

    I can kind of see how this would be true, but I am having a lot of trouble formalizing a proof. If someone could help me along that would be great. I have just started this course, and it's a little difficult to get used to.
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    Re: Simple Topology Proof

    Quote Originally Posted by Aryth View Post
    Suppose U and V are pointsets of \mathbb{R}. If U and V both have a leftmost point, prove that the union U \cup V also has a leftmost point.
    Suppose that \alpha is the left most point of U. That means that \alpha\in U and \forall x\in U we have \alpha\le x~..
    Make similar about \beta the left most point of V.

    Clearly both \alpha~\&~\beta belong to U\cup V.

    Can you finish?
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    Super Member Aryth's Avatar
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    Re: Simple Topology Proof

    Hmmm... So... Just do a similar construction?

    Since \alpha \ \& \ \beta belong to U \cup V, then \forall x \in U \cup V either \alpha \leq x or \beta \leq x, which guarantees that there is a leftmost point in U \cup V.
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    Re: Simple Topology Proof

    Quote Originally Posted by Aryth View Post
    Hmmm... So... Just do a similar construction?

    Since \alpha \ \& \ \beta belong to U \cup V, then \forall x \in U \cup V either \alpha \leq x or \beta \leq x, which guarantees that there is a leftmost point in U \cup V.
    No you are not done.
    There is only one left most point of U\cup V.

    Now it is one of \alpha\text{ or }\beta. Now we know that \[\left( {\alpha  = \beta {\kern 1pt} } \right){\kern 1pt}  \vee \left( {{\kern 1pt} \alpha  < \beta } \right){\kern 1pt} {\kern 1pt}  \vee \left( {{\kern 1pt} \alpha  < \beta } \right){\kern 1pt} {\kern 1pt} \].

    Finish it!
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    Super Member Aryth's Avatar
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    Re: Simple Topology Proof

    I'm slightly confused... I know what you're saying, but I can't see the rest of the proof... I mean, I could suppose that \alpha \leq \beta and that would make \alpha the left most point, or I could say the opposite, and \beta would be the left most point. I feel like I'm missing something.
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    Re: Simple Topology Proof

    Quote Originally Posted by Aryth View Post
    I'm slightly confused... I know what you're saying, but I can't see the rest of the proof... I mean, I could suppose that \alpha \leq \beta and that would make \alpha the left most point, or I could say the opposite, and \beta would be the left most point. I feel like I'm missing something.
    You know \[\left( {\alpha  = \beta {\kern 1pt} } \right){\kern 1pt}  \vee \left( {{\kern 1pt} \alpha  < \beta } \right){\kern 1pt} {\kern 1pt}  \vee \left( {{\kern 1pt} \alpha  < \beta } \right){\kern 1pt} {\kern 1pt} \].

    Take each case. Prove that \alpha\text{ or }\beta is the left most point..
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    Super Member Aryth's Avatar
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    Re: Simple Topology Proof

    Ok...

    Note: If x \in U \cup V, then \left[x \in U \vee x \in V \vee (x \in U \wedge x \in V)\right]. In any of these cases, either \alpha or \beta is the left most point.

    We know: \[\left( {\alpha  = \beta {\kern 1pt} } \right){\kern 1pt}  \vee \left( {{\kern 1pt} \alpha  < \beta } \right){\kern 1pt} {\kern 1pt}  \vee \left( {{\kern 1pt} \beta  < \alpha } \right){\kern 1pt} {\kern 1pt} \]

    Let \alpha = \beta, then they are the same point, and that point is the left most point, since it is also the left most point of both U and V.

    Let \alpha < \beta, then \forall x \in U and \forall x \in V, \alpha \leq x. In other words, it means that \alpha is to the left of all points in V, and is therefore the left most point of the union.

    Let \beta < \alpha , then \forall x \in U and \forall x \in V, \beta \leq x. In other words, it means \beta is to the left of all points in U, and is therefore the left most point of the union.
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    Re: Simple Topology Proof

    Well done. It is now complete.
    I think it is generally true that topologist are sticklers for completeness in proofs .
    Thanks from Aryth
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    Re: Simple Topology Proof

    I really appreciate the help. And I'll keep that in mind, I'm getting that feeling from my professor.
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