# Problems with Topology question

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• Dec 30th 2008, 01:47 PM
HTale
Problems with Topology question
I'm stuck on a question. It states to consider the set $\displaystyle Y = [-1,1]$ as a subspace of $\displaystyle \mathbb{R}$, it being the standard topology. And now I have to consider which are open in $\displaystyle \mathbb{R}$ and which are open in $\displaystyle Y$.

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

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

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

Thanks in advance.
• Dec 30th 2008, 01:58 PM
ThePerfectHacker
Quote:

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

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

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

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

Thanks in advance.

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

Now, $\displaystyle A = (-1,-1/2) \cup (1/2,1)$. The set $\displaystyle (-1,-1/2) = (-1,-1/2)\cap [-1,1]$ and $\displaystyle (1/2,1) = (1/2,1)\cap [-1,1]$. Thus, these are open sets and so the union of two open sets is open.
• Dec 30th 2008, 02:55 PM
Plato
Quote:

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

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

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

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

Set $\displaystyle D$ is closed in $\displaystyle \mathbb{R}$, and it is closed in $\displaystyle \mathbb{Y}$.
Can you explain this?
• Dec 30th 2008, 08:54 PM
HTale
Quote:

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

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

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

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

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