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

**Mollier** · January 3rd 2011, 10:56 PM

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.

**Drexel28** · January 3rd 2011, 11:12 PM

Originally Posted by Mollier

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?

**Mollier** · January 3rd 2011, 11:22 PM

Great explanation, thanks!

**DrSteve** · January 4th 2011, 03:07 AM

Drexel. I think you mean "The reason it's not open" in your post.

In any case, it seems that the post was understood.

**Mollier** · January 4th 2011, 03:27 AM

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.

**TheQuantiser** · July 28th 2012, 04:15 PM

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?

**Deveno** · July 28th 2012, 10:43 PM

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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