# Problems with Topology question

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• Dec 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.
• Dec 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.
• Dec 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?
• Dec 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.