What does it mean for a set to be open inside another set? I am taking an undergraduate analysis course, and last week we began to deal with connectedness. Before defining connectnedness, our instructor gave us an alterative definition of an open set. This is the definition:
Let ${U\subset S\subseteq \mathbb{R}^n}$. ${U}$ is open in ${S}$ if ${U = S\cap V}$, for some ${V}$ open in ${\mathbb{R}^n}$.
I don't understand for what it means for the set ${U}$ to be open in ${S}$, as opposed to it being open in ${\mathbb{R}^n}$. I'm trying to correlate this with the definition of an open set as given in Mathematical Analysis by Apostol, which is that a set ${A \subseteq \mathbb{R}^n}$ is open if ${\forall}$ ${x \in A}$, ${\exists}$ ${\epsilon > 0}$ ${|}$ ${B_\epsilon(x) \subseteq A}$. 
Does it even make sense to correlate the two? Or is openness inside a set different from openness in general, and this new definition should be accepted as gospel? 
 A: The general problem is this: You have a set $X$ (in your case, $\mathbb R^n$), and you've defined a topology on that set, which essentially means, you've defined what it means for a set to be open in $X$ ($\mathbb R^n$). Note that a topology is always defined over some specific set (one of the rules is, the set itself is always open).
Now consider a subset $S$. You want to do topology on that subset, so you need to say what are the open sets on this subset.
Now, the obvious idea would be that a set is open in $S$ iff it is open in $X$. Unfortunately, this only works if $S$ happens to be open in $X$, but you want to do topology also on other subsets. For example, you might consider a straight line (a copy of $\mathbb R$ in $\mathbb R^n$; indeed, in that case you want the topology to be exactly the usual topology of $\mathbb R$, so all the results on $\mathbb R$ that you know are still valid.
But if just taking the open sets as is doesn't always work, what can we do then?
Well, if the open sets don't fit, we just make them to fit. What we do is to just cut away those parts of the open sets $V$ in $X$ ($\mathbb R^n$) that are outside $S$ (that's exactly what $S \cap V$ does), and declare the resulting sets the open sets of $S$. And it turns out that this works perfecly:


*

*It always gives a valid topology, no matter which subset $S$ we consider.

*Open sets in $X$ that already happen to be subsets of $S$ are still open as subsets of $S$.

*If $S$ happens to be open in $X$, we get the very same result as the first idea.

*If $X$ is a metric space (like $\mathbb R^n$) with the topology defined by its metric, then you get the same topology this way as you get when you first restrict the metric to $S$ (making $S$ a metric space in its own right), and then use that metric to define the topology of $S$. Note that this includes, but is not limited to, the example of straight lines in $\mathbb R^n$.
A: That's not an "alternative" definition of open set. It's the definition of the so-called subspace topology. It is the best way to define a topology (i.e. which sets are considered open) on $S$ which agrees with the topology of $\Bbb R^n$.
For instance, in $S=[0,1]\subseteq \Bbb R$ (where $\Bbb R$ has the standard topology), we have that $[0,\frac12)$ is considered open, because you can write it as the intersection between $S$ and some open interval, like $(-1,\frac12)$. Of course, something like $(\frac13,\frac23)$ is also open in $S$.
And yes, it makes some sense to correlate the two. A subset $U\subseteq S$ is open iff for any $s\in S$ there is an $\epsilon$ such that $B_\epsilon(s)\cap S\subseteq U$. Or you could consider $S$ as a metric space in its own right (with the metric inherited from $\Bbb R^n$), and in that case you can actually remove the "${}\cap S$" from the above (because the interpretation of $B$ changes), and it will truly look exactly the way your used to.
