Can I define the compactness using the closure or boundary? I've seen that I can define a topological space and continuous function with the closure or boundary without the open set. Moreover, there are definitions of the connectedness:


*

*If $X=X_1\cup X_2$ and $X_1\cap\operatorname{cl}(X_2)=\operatorname{cl}(X_1)\cap X_2=\varnothing$, then $X_1$ or $X_2$ is the empty set and $X_1\neq X_2$.

*The only subsets of $X$ with empty boundary are $X$ and the empty set.


What about the compactness? Should I use the finite intersection property? 
 A: Some characterisations of compactness that use closedness/closure:
$X$ is compact iff for every space $Y$ the projection $\pi_Y: X \times Y \to Y$ is a closed map (so $\pi_Y[A]$ is closed whenever $A \subseteq X \times Y$ is). But this is sort of "external" to $X$, but elegant nonetheless.
$X$ is compact iff for every family of subsets $\mathcal{F}$ of $X$ that has the finite intersection property, we have $\bigcap \{\overline{F}: F \in \mathcal{F}\} \neq \emptyset$. We can also restrict ourselves to consider only families of closed subsets that have the FIP, if you prefer. 
I cannot think of any that use the boundary, though.
A: Yes, you can define that $K \subseteq X$ is compact if for every family $\mathcal{F}$ of closed sets in $K$ satisfying the finite intersection property, we have $\bigcap \mathcal{F} \ne \emptyset$. 
Alternatively, $K$ is compact if for every family of closed sets in $K$ satisfying $\bigcap \mathcal{F} = \emptyset$ there exists a finite subset of $\mathcal{F}$ with empty intersection.
