# A simple question about open set

Let $X=\mathbb{R}$ and $S=[-1,1]$, i don't quite understand why $[-1,0)$ is open in $S$ but not open in $X$. As far as i know, to show a set is open, we need to show there exist an open ball $B(x,r), r> 0$ st for all $x\in [-1,0)$ ,$x\in B\subset S$, however, i still can't see why $[-1,0)$ is open in $S$ but not open in $X$. When we look at -1, i don't think this is a interior point of S.

-
But what does it mean for something to be open *in $S$* under the subspace topology that $S$ gains as a subset of $\mathbf R$? –  Dylan Moreland Jun 19 '12 at 7:10

The open balls in $S$ are different than the open balls in $X$. For example, we have an open ball around $-1$ in $S$ contained in $[-1,0)$, that is $B(-1,1)$. But the open ball $B(-1,1)$ in $X$ is not contained in $[-1,0)$, as it is equal to $(-2,0)$.

In general, given any space $X$ and a subset $S$ you have sets which are open in $S$ but not in $X$. This is because the open sets in $S$ are of the form $S\cap U$ where $U\subseteq X$ is open in $X$, and not all sets of the form $S\cap U$ are open in $X$.

-
isn't -1 is the limit point ? –  Mathematics Jun 19 '12 at 7:19
I'm not sure what you mean. $-1$ is a limit point of $S$ and $X$, but I'm not sure why that is relevant here. –  Alex Becker Jun 19 '12 at 7:21
@Mathematics: $B(x,r)$ is defined as a subset of the space you are working in. How can $B(-1,r)$ not be a subset of $S$? –  Erick Wong Jun 19 '12 at 7:26
@Mathematics $B(-1,r)=\{x\in S:d(x,-1)<r\}$ as a nhood of $-1$ in $S$ is by definition a subset of $S$. Ofcourse the open ball in $X$ is not a subset of $S$. –  Thomas E. Jun 19 '12 at 7:26
@Mathematics The problem here is one of notation. The notation $B(-1,r)$ is not specific enough. You must say whether you are talking about $B(-1,r)$ in $S$ or in $X$. –  Alex Becker Jun 19 '12 at 7:28

To show $[-1, 0)$ is open in $[-1,1]$, it is easier to use some general topology.

Suppose $A \subset X$. $U \subset A$ is open in the subset or relative topology on $A$ if and only if there exists $V \subset X$ open in $X$ such that $U = A \cap U$.

You can also prove this using the open ball definition, but now open balls in $[-1,1]$ is just the open ball in $\mathbb{R}$ intersected with $[-1,1]$, so a proof using the open definition essentially reduces to the above.

Let $A = [-1,1]$ and $X = \mathbb{R}$. $[-1, 0) = [-1,1] \cap (-2, 0)$. $(-2, 0)$ is open in $\mathbb{R}$. So by definition $(-1, 0)$ is open in $[-1,1]$.

To show that $[-1,0)$ is not open in $\mathbb{R}$, you need to consider what the actually topology on $\mathbb{R}$. As you mentioned, you can consider the open ball definition. If $[-1,0)$ was open, then for any $x \in [-1, 0)$, there exists a ball centered at $x$ entirely inside $[-1,0)$. Clearly no ball centered at $-1$ can have this property. Thus $[-1,0)$ is not open.

-

$[-1,0)$ itself is a neighborhood of $-1$ in $S$, so $-1$ is an interior point.

EDIT: Another way to think about it: the complement of $[-1,0)$ relative to $S$ is $[0,1]$, which is closed. But the complement relative to $\mathbb{R}$ is not closed.

-
similar to the comment to Alex, isn't -1 is the limit point? –  Mathematics Jun 19 '12 at 7:19
Yes. But it is also an interior point. –  Andrew Salmon Jun 19 '12 at 7:34
The only reason why $-1$ is a limit point is because it lies in the set @Mathematics. –  Don Larynx Nov 6 '13 at 3:39
That's because $-1$ is the left bound of the space. That is, $[-1, 0)$ is relatively open to $S$ because $[-1, 0) \subseteq G \cap [-1, -1]$ where $G$ is a union of open sets in $X$ such that $r = 1$ is a bound for $G$.