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Math Help - Segment (a,b), subset of R^2, closed, open (from Baby Rudin)

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    Member Mollier's Avatar
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    Segment (a,b), subset of R^2, closed, open (from Baby Rudin)

    Hi,

    Rudin defines the segment (a,b) to be the set of all real numbers x such that a<x<b.
    In one example he considers the subset (a,b) of R^2 and says that it is not an open set if we regard it as a subset of R^2, but that it is open if we regard it as a subset of R^1.

    How do I regard the segment (a,b) as a subset of R^2? The way I see it from his definition of a segment, it is just a part of the "x-axis" and would be open just as in R^1...

    Thanks.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Mollier View Post
    Hi,

    Rudin defines the segment (a,b) to be the set of all real numbers x such that a<x<b.
    In one example he considers the subset (a,b) of R^2 and says that it is not an open set if we regard it as a subset of R^2, but that it is open if we regard it as a subset of R^1.

    How do I regard the segment (a,b) as a subset of R^2? The way I see it from his definition of a segment, it is just a part of the "x-axis" and would be open just as in R^1...

    Thanks.
    What he really means is that you can think of it as (a,b)\times\{0\}\subseteq\mathbb{R}^2. The reason it's open is if you take \displaystyle \frac{a+b}{2} every neighborhood of that point will hit "non-x-axis" stuff which isn't in (a,b) and thus \displaystyle \frac{a+b}{2} is not an interior poitn of (a,b) and so (a,b) can't be open?
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    Member Mollier's Avatar
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    Great explanation, thanks!
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    Drexel. I think you mean "The reason it's not open" in your post.

    In any case, it seems that the post was understood.
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    Member Mollier's Avatar
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    Quote Originally Posted by DrSteve View Post
    Drexel. I think you mean "The reason it's not open" in your post.

    In any case, it seems that the post was understood.
    Yeah, I figured it was a typo, but thanks anyway.
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    Re: Segment (a,b), subset of R^2, closed, open (from Baby Rudin)

    This is a pretty old thread, but I'm taking the liberty of raising an unresolved issue with the original question.

    While an explanation for (a,b) being non-open in R^2 has been suggested, it does not explain how R^1 may be regarded as a subset of R^2, as Rudin implies in pg. 35 "Example 2.2(g) showed that a set may be open relative to Y without being an open subset of X." In short, how is the interval open and non-open in the same metric space?
    Last edited by TheQuantiser; July 28th 2012 at 04:23 PM.
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    Re: Segment (a,b), subset of R^2, closed, open (from Baby Rudin)

    in general, the way we regard \mathbb{R} as a subset of \mathbb{R}^2 is to identify it with the set \mathbb{R} \times \{0\}.

    this sends the point x \in \mathbb{R} \to (x,0) \in \mathbb{R}^2 (the real line becomes "the x-axis").

    the non-null intersection of an open disk in \mathbb{R}^2 with \mathbb{R} \times \{0\}, is just an open interval on the x-axis.

    these "open intervals on the x-axis" are open in the relative topology on the x-axis induced by the topology on the plane, but are not open in the full plane.
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