Understanding the Category of open subsets of a top. space X $Op_X$ I have a problem very similar to the one posted here by Brian. I guess he also stumbled across the introduction of the category of open subsets of a topological space X (denoted by $Op_X$) in Pierre Schapira's lecture notes on Algebra and Topology (or one of his sources). There you can read:

Let $X$ be a topological space. The family of open subsets of $X$ is ordered by inclusion and we denote by $Op_X$ the associated category. Hence: $$\mathrm{Hom}_{Op_X}(U,V) =\begin{cases} \{\text{pt}\} & \text{if }U \subset V,\\ 
\emptyset & \text{otherwise.}
\end{cases}$$

Okay, $\{\text{pt}\}$ is simply defined beforehand as a set with one single element, I get that. $U$ and $V$ are open subsets of $X$ obviously, that's okay with me, too.
What I don't get is:


*

*Why "hence" - isn't that pure definition? Or is it more like "so the only possible/useful/... definition of the $\text{Hom}$-sets is: [...]"?

*So, if $U \subset V$, we have one morphism $pt:U\to V$. I feel like everybody knows which morphism this is, but I don't see it. What does $pt$ do? Is this the identity-morphism, namely $pt(u)=u ~~\forall u\in U$ (which would include $U$ in $V$ if $U\subset V$, which would make sense somehow)? If my guess is wrong, then we must at least have some kind of unique morphism, which we can always find if $U\subset V$. But how do we get that one (or its existence)? (I somehow feel like this has got something to do with fhyve's argument about isomorphisms mentioned in the question linked above. I think I understood why we have isomorphisms in that situation, but I don't see a connection to why $Op_X$ is a category or to why/how we have that $pt$-morphism. Also another user states that $pt$ is the inclusion map. Is that correct and is the inclusion map the same as what I guessed a few lines above?)


I am new to M.SE and also relatively new to category theory, so please be kind. (: I hope my English is okay.
(I'm not sure if I should have posted this question directly to Brian's question - as I would have at a standard online board - but regarding this link from the meta, I guess this is the right way.)
I hope you guys can help me out, any effort would be much appreciated!
//Edit: As a final note, I wanted to say thank you to all contributors. I will now accept one answer (the correct/shortest answer for me would be the one about inclusion markers, but I'll accept the one that really helped me understanding what's going on with this category, hope everybody is okay with that. All of the answers and comments were really helpful and of very high quality.).
 A: If you have a partially order set (a poset) $X$, then you can define the category $\cal X$ whose objects are the elements in $X$ and whose morphisms are given by
$$\mathrm{Hom}_{\cal X}(x,y)=
\begin{cases}
\{x\to y\}, & \text{if }x \le y\\ 
\emptyset,  & \text{otherwise}
\end{cases}$$
One can also define it the other way round, with a morphism $y\to x$ if $y\ge x$. It is just a matter of taste. Some people like to have morphisms going "up", other like them going "down". Of course each definition gives the opposite category of the other definition.
The family of open subsets of a topological space is a poset with $U\le V$ if $U\subseteq V$. That is why the associated category (the one induced by the order) is precisely the category descirbed in your post.
But note that in this particular case we can indeed think of the morphisms as inclusions. We could even define the category to consist of all the inclusion maps, since an inclusion $U\hookrightarrow V$ exists if and only if $U\subseteq V$.
A: The answers by Stefan and Eric make good points, and I would like to add to them.
It is common to describe the notion of a category abstractly as a collection of labeled dots and arrows.  This is rather like describing a group by generators and relations.  You may ask "what do the morphisms do?'' but this question is essentially meaningless at this point.  Just like in group theory, the group elements don't "do" anything until you pick a representation $\rho:G\rightarrow \operatorname{Aut}(V)$ of this group as a collection of linear operators on maybe some vector space.
In order to relate to your example, note first off that there are two categories in play here.  Firstly, there's $Op_X$ that you defined in your question.  Secondly, there's the category $Op_X'$ with the same set of objects (open sets of $X$) and morphisms given by the inclusion maps.  It might be helpful to think of $Op_X$ as the abstract category and $Op_X'$ as a particular representation of $Op_X$.  In this case, there is an obvious isomorphism $Op_X\cong Op_X'$, so you'd be excused for identifying them in your mind.  But in many cases it is better to distinguish between the "abstract" object and its "concrete" realization, whatever this may mean.
A: I think the answer is that morphisms do not have to correspond to functions, and in this case pt does not need to correspond to any function - it's just kind of a marker which tells you about an inclusion.
