Let $X = [0, + \infty[ \subset \mathbb{R}$ and $\tau$ be a topology on $X$ which is composed by $X$, $\emptyset$ and every set of the form $]a ; + \infty[$, where $a \in \mathbb{R}$.
Can somebody explain to me why $(X, \tau)$ is a compact topological space ?
If I take the family of set $(U_{i})_{i \in \mathbb{N}}$ such that for every $i \in \mathbb{N}$, $U_{i} = ]\frac{1}{n}, + \infty[$, then we have that $\bigcup_{i \in \mathbb{N}} U_{i} = X$ (so $(U_{i})_{i \in \mathbb{N}}$ is an open cover of $X$), but $(U_{i})_{i \in \mathbb{N}}$ has no finit subcover of $X$ right (or I'm probably missing something) ?
Thank you for your help.
The union of your sets $U_i$ is only the interval $(0, \infty)$. To show that $X$ is compact, take some open cover $U_i$ with $\bigcup U_i = X$. Since $0 \in X$, there is some $U_j$ with $0 \in U_j$. But then $U_j = X = [0, \infty)$ and thus $\{U_j\}$ is a finite subcover.