I have a doubt at the definition of compact spaces. So if you have a topological space $X$, then $X$ is compact if every open cover of $X$ has a finite subcover. In other words, if $X$ is the union of a family of open sets, there is a finite subfamily whose union is $X$.
So, if $\{A_i\}$ is a family of open sets such that it is a finite subcover of $X$ so $X=$ union of all $A_i$. Ok, until here I got it. But now if I consider that there is an $j$ such that $A_j=X$, which is open, so of course that we have ALWAYS: $X$=union of all $A_i$.
My doubt: this might be stupid but if I can do this always so the set is always compact...??
Thanks in advance to the ones who help me :D :D
(Source of definition: http://mathworld.wolfram.com/CompactSpace.html)
No, the definition says:
This means that you cannot select the sets $A_j$. You are given a set $X$, and a set of sets $\{A_j\}_{j\in J}$. Then, no matter what the sets $A_j$ are (you do not know if one of them is equal to $X$), you can find a finite set $A_1,A_2,\dots, A_n$ such that they cover $X$.
Sure, there may be (and in fact always are) covers of $X$ which do contain finite subcovers. A typical example of this is taking any cover that contains $X$ as one of its covering elements, as then $\{X\}$ is obviously a finite subcover. But that's not the point. The point is that there must be a finite subcover no matter what the original cover was.
For example, $X=\mathbb R$ is not compact, because I can define
$$A_i = (i-1, i+1)$$
and then $\{A_i, i\in \mathbb Z\}$ becomes a cover for $\mathbb R$. However, there is no finite subcover that covers $\mathbb R$.