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

• Jan 3rd 2011, 10:56 PM
Mollier
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.
• Jan 3rd 2011, 11:12 PM
Drexel28
Quote:

Originally Posted by Mollier
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?
• Jan 3rd 2011, 11:22 PM
Mollier
Great explanation, thanks!
• Jan 4th 2011, 03:07 AM
DrSteve
Drexel. I think you mean "The reason it's not open" in your post.

In any case, it seems that the post was understood.
• Jan 4th 2011, 03:27 AM
Mollier
Quote:

Originally Posted by DrSteve
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.
• Jul 28th 2012, 04:15 PM
TheQuantiser
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?
• Jul 28th 2012, 10:43 PM
Deveno
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.