Problems with Topology question

Printable View

December 30th 2008, 02:47 PM
HTale

Problems with Topology question

I'm stuck on a question. It states to consider the set $Y = [-1,1]$ as a subspace of $\mathbb{R}$ , it being the standard topology. And now I have to consider which are open in $\mathbb{R}$ and which are open in $Y$ .

The first set is $A = \{x : 1/2 < |x| < 1\}$ , which I think is $(1/2, 1) \cup (-1, -1/2)$ . Now, that is open in $\mathbb{R}$ , because it is the union of two open sets in $\mathbb{R}$ . But is it open in $Y$ ? How do I tell if it is? How about the following cases:

$(1/2, 1] \cup [-1, -1/2)$
$[1/2, 1) \cup (-1, -1/2]$
$[1/2, 1] \cup [-1, -1/2]$

My guess is that they are all closed in $\mathbb{R}$ , but all open in $Y$ . Am I right? Why/why not?

Thanks in advance.
December 30th 2008, 02:58 PM
ThePerfectHacker

Quote:

Originally Posted by HTale

I'm stuck on a question. It states to consider the set $Y = [-1,1]$ as a subspace of $\mathbb{R}$ , it being the standard topology. And now I have to consider which are open in $\mathbb{R}$ and which are open in $Y$ .

The first set is $A = \{x : 1/2 < |x| < 1\}$ , which I think is $(1/2, 1) \cup (-1, -1/2)$ . Now, that is open in $\mathbb{R}$ , because it is the union of two open sets in $\mathbb{R}$ . But is it open in $Y$ ? How do I tell if it is? How about the following cases:

$(1/2, 1] \cup [-1, -1/2)$
$[1/2, 1) \cup (-1, -1/2]$
$[1/2, 1] \cup [-1, -1/2]$

My guess is that they are all closed in $\mathbb{R}$ , but all open in $Y$ . Am I right? Why/why not?

Thanks in advance.

A subset $X$ is open in $[-1,1]$ iff $X = Y \cap [-1,1]$ where $Y$ is open in $\mathbb{R}$ .

Now, $A = (-1,-1/2) \cup (1/2,1)$ . The set $(-1,-1/2) = (-1,-1/2)\cap [-1,1]$ and $(1/2,1) = (1/2,1)\cap [-1,1]$ . Thus, these are open sets and so the union of two open sets is open.
December 30th 2008, 03:55 PM
Plato

Quote:

Originally Posted by HTale

It states to consider the set $Y = [-1,1]$ as a subspace of $\mathbb{R}$ , it being the standard topology. And now I have to consider which are open in $\mathbb{R}$ and which are open in $Y$ .
$B=(1/2, 1] \cup [-1, -1/2)$
$C=[1/2, 1) \cup (-1, -1/2]$
$D=[1/2, 1] \cup [-1, -1/2]$
My guess is that they are all closed in $\mathbb{R}$ , but all open in $Y$ . Am I right? Why/why not?

By “being the standard topology”, I assume you the relative topology on $\mathbb{Y}=[-1,1]$ .
I added letters to your set for reference.

Set $B$ is neither open nor closed in $\mathbb{R}$ , however it is open in $\mathbb{Y}$ . Can you explain this?

Set $C$ is neither open nor closed in $\mathbb{R}$ . It is not open in $\mathbb{Y}$ because it contains a boundary point $\frac{1}{2}$ and it is not closed because $-1$ is a limit point not in the set.

Set $D$ is closed in $\mathbb{R}$ , and it is closed in $\mathbb{Y}$ .
Can you explain this?
December 30th 2008, 09:54 PM
HTale

Quote:

Originally Posted by Plato

By “being the standard topology”, I assume you the relative topology on $\mathbb{Y}=[-1,1]$ .
I added letters to your set for reference.

Set $B$ is neither open nor closed in $\mathbb{R}$ , however it is open in $\mathbb{Y}$ . Can you explain this?

Set $C$ is neither open nor closed in $\mathbb{R}$ . It is not open in $\mathbb{Y}$ because it contains a boundary point $\frac{1}{2}$ and it is not closed because $-1$ is a limit point not in the set.

Set $D$ is closed in $\mathbb{R}$ , and it is closed in $\mathbb{Y}$ .
Can you explain this?

Thank you very much, I can explain all of them now.