What does it really mean "take a set from the topology $\tau$"? It's open or closed?

Question

I've started my topology classes like one month ago, so please excuse me if this is a silly question to ask, but I need to understand it to move forward.

I have an exercise that states:

Let $(X,\tau)$ be a topological space. Consider $Y=U\sqcup\{a\}$, and $\tau'=\{X\sqcup \{a\}| X\in\tau\}$. Prove that $(Y,\tau')$ is a topology.

Okay, so here on $\tau'$ we're taking an open set of $\tau$. That's what they want us to understand from that $X\in\tau$.

But two exercises below that, we have another one:

Let $A$ and $B$ subspaces of a topological space $X$ (that is, $A$ and $B \subset (X,\tau)$). Prove the following

$i)$ $Int(Int(A))=Int(A)$, $Int(A\cap B)=Int(A)\cap Int(B), \dots$

And more equalities that I'm asked to prove (7 more), but that's not the point on here. What I can understand now is that they want me to take two arbitrary subsets, open or closed.

So what does it mean to take a set from a topology? It's always open? Or can be open or closed?

Sorry if I didn't explained it well, I'll answer further question if something can't be understood.

Thanks for your time.

Answer 1

If $(X, \tau)$ is a topological space, then the elements of $\tau$ are precisely the open sets. Thus, $A \in \tau$ means that $A$ is open. Note that your book uses really sloppy notation: the space $(X, \tau)$ uses, the letter $X$, but later the letter $X$ is also used to refer to an arbitrary element of $\tau$, in the definition of $\tau'$. (Also, the definition of $\tau'$ is technically wrong: as written, $\tau'$ does not contain the empty set, which is a requirement for $\tau'$ to be a topology.)

A subset (or subspace) of $X$, without qualification, can be subset without restrictions given by the topology of $X$. It can be open, it can be closed; but it can also be neither or both! Don't make the mistake of thinking that "closed" means "not open", despite the initially confusing terminology. A topological space always has a set that is open and closed, namely the entire space itself; and most topological spaces have subsets that are neither open nor closed (for example, the rational numbers $\mathbb Q \subseteq \mathbb R$ with the usual topology).

Answer 2

No, this is not about taking a set from a topology. It's about taking a subset from a set. Consider the reals, for instance, with its usual topology. Given any subset of $\mathbb R$, is it true or false that $\overline A=\overline{\overline A}$? It's just this. On order to consider a subset $A$ of $\mathbb R$, you don't need to see it as a topological space. It's just a set. The topolgy only becomes relevant when you try to determine whether $\overline A=\overline{\overline A}$ or not.