Specific Question About Open/Close Sets. So I had a question about open/close subsets because we started this in Topology today.
So let's take an closed subset of $\mathbb{R}$ so, for example $X = [-2, 2]$.
Let's say I want to compare if a set is closed in $X$ vs a set closed in $\mathbb{R}$ what would be the main difference?
So let's say I have 3 sets, say $Y_1 = \{x\ | 1 < x < 2 \}$, $Y_2 = \{x\ | 1 \leq x < 2 \}$, $Y_3 = \{x\ | 1 < x \leq 2 \}$.
Like looking at $Y_1$ this set is open in $X$ because $1$ and $2$ are limit points of the set and they're not contained in $Y_1$ and similarly to $\mathbb{R}$.
So for $Y_2$ and $Y_3$ would these not be open in $X$ and $\mathbb{R}$? Is there a difference between comparing closed subsets of $\mathbb{R}$ such as $X$ vs $\mathbb{R}$ given the sets I have defined? 
 A: $Y_1$ is indeed open, in both $X$ and $\mathbb{R}$, but not for the reason you say. It's open, because every point is an interior point, or you could look at the complement, which in $\mathbb{R}$ is $(-\infty,1] \cup [2, \infty)$, which is indeed closed (it contains all its limit points). But openness is more direct, as $Y_1$ is an open interval which is open by definition (if you use the order topology definition) or an open ball (around $1\frac{1}{2}$ with radius $\frac{1}{2}$) if you use the metric definition, so directly open in $\mathbb{R}$. 
If $A \subset X \subset \mathbb{R}$, and $A$ is open in $\mathbb{R}$, then $A$ is also open in $X$. In general, $A$ is open in $X$ iff there exists some $A'$ open in $\mathbb{R}$ such that $A' \cap X = A$. Here we can use $A' = A$, e.g. for $Y_1$.
The same holds for closed sets as well. 
But a set can be open in $X$ without being open in $\mathbb{R}$, and $Y_3$ is an example of that. We can take for $A = Y_3$ the set $A' = \{1 < x < 3 \}$, which is open in the reals (open interval etc.) and $A' \cap X = A$. But $Y_3$ is not open in $\mathbb{R}$, as $2$ is not an interior point of $Y_3$ in $\mathbb{R}$ (but it is an interior point in $X$!).
The set $Y_2$ is neither open in $\mathbb{R}$ not in $X$, now failing both at $x=1$ to be an interior point.
A: Because $X$ is closed in $\mathbb R$ all closed subsets of $X$ are also closed in $\mathbb R$.
Proof: Take $A \subseteq X$ close. Thus $X\setminus A$ is open. Also $\mathbb R\setminus X$ is open because $X$ is close in $\mathbb R$. So $\mathbb R \setminus A = (\mathbb R\setminus X) \cup (X\setminus A)$ is open which proves that $A$ is close in $\mathbb R$.
The same is not true, if $X$ is not closed in $\mathbb R$. Take for example $\tilde X = ]-2,2[$. Regarding to $\tilde X$ the set $Y_2 = \{x|1\le x<2\}$ is closed but $Y_2$ is not closed in $\mathbb R$.
Note: Not all open subsets of $X$ are open in $\mathbb R$ as you can see in the comments...
