Ambiguity in definition of compactness I am struggling with the definition of compactness in a topological sense. Below is the definition presented in my lecture notes:

A topological space $X$ is compact if every open cover has a finite
  subcover on $X$.

Okay, this seems to make sense. But, an example is later presented in the notes: $X$ with the trivial topology $\tau = \{X, \varnothing \}$ is compact. Again, okay, this seems to make sense, as any open cover is finite, if we look at $\tau$. 
My question is: Does the cover come from the set $X$ or $\tau$? Apologies if my thought process seems unclear!
 A: open cover (for a given topology $\tau$): a collection of open sets $U_{\alpha}$ (open with respect to the topology $\tau$) such that $$X \subset \underset{\alpha \in A}{\bigcup} U_{\alpha}$$
open set: A set in the topology $\tau$ of the space $X$.
compact (for a given topology $\tau$): every open cover of $X$ (with respect to the topology $\tau$) has a finite subcover.
A given open cover therefore depends on both:


*

*the given space $X$

*the topology $\tau$ given for $X$


If $\nu$ is a different topology for $X$, then $(X,\tau)$ and $(X,\nu)$ are different topological spaces. Therefore, which collections of sets are open covers of $X$ and as a result whether or not $X$ is compact will be different.
A: You should use elements of the topology (open sets) to cover your space.
A: Also in your lecture notes should be:
Let $X$ be a set, $(X,\mathcal{T})$ a topological space. Then a collection $\mathcal{A} \subset \mathcal{P}(X)$ is called an open cover if:
(i) $\mathcal{A}$ is a cover, i.e. $\forall x \in X, \exists A \in \mathcal{A}$ such that $x \in A$.
(ii) Elements of $\mathcal{A}$ are open, i.e. $\mathcal{A} \subset \mathcal{T}$.
So the cover most definitely comes from $\mathcal{T}$.
A: An open cover is a collection of open sets whose union is $X$. Thus it is a subset of $\tau$ to begin with.
Since every subset of $\tau$ is finite, then so is every open cover.
