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

    How do I regard the segment $\displaystyle (a,b)$ as a subset of $\displaystyle R^2$? 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 $\displaystyle R^1$...

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

    How do I regard the segment $\displaystyle (a,b)$ as a subset of $\displaystyle R^2$? 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 $\displaystyle R^1$...

    Thanks.
    What he really means is that you can think of it as $\displaystyle (a,b)\times\{0\}\subseteq\mathbb{R}^2$. The reason it's open is if you take $\displaystyle \displaystyle \frac{a+b}{2}$ every neighborhood of that point will hit "non-x-axis" stuff which isn't in $\displaystyle (a,b)$ and thus $\displaystyle \displaystyle \frac{a+b}{2}$ is not an interior poitn of $\displaystyle (a,b)$ and so $\displaystyle (a,b)$ 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 $\displaystyle (a,b)$ being non-open in $\displaystyle R^2$ has been suggested, it does not explain how $\displaystyle R^1$ may be regarded as a subset of $\displaystyle R^2$, 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; Jul 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 $\displaystyle \mathbb{R}$ as a subset of $\displaystyle \mathbb{R}^2$ is to identify it with the set $\displaystyle \mathbb{R} \times \{0\}$.

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

    the non-null intersection of an open disk in $\displaystyle \mathbb{R}^2$ with $\displaystyle \mathbb{R} \times \{0\}$, 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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