# Proving [0,1] is not compact with a particular topology

The background of my question is $$(\mathbb{R},\mathcal{F})$$, the real line with $$\mathcal{F}$$ a topology defined as $$\mathcal{F}:=\{\emptyset, \{[x,t) : t > x\}\}.$$

Now, how is the prove to get $$[0,1]$$ is nos compact? If it is possible, I´d like to see a simple prove (that is, not using sequences).

For example: $$\{[0,\frac{1}{2}),[\frac{1}{2},2)\}$$ is a covering of $$[0,1]$$ such that not admitts a subcovering?

Thank you very much, and kind regards!

• See the properties of this – Dog_69 Jan 16 at 16:15

Remark that $$\{1\}$$ is open in $$[0,1]$$ for the induced topology since $$\{1\}=[1,2)\cap [0,1]$$. Consider $$[0,1-1/n)\cup\{1\}$$ its a covering which does not has a finite subcovering.
• @PeterSzilas $\{1\}$ is an open set of $[0,1]$ in this topology as Tsemo already stated. So the open cover is $\{[0,1-\frac1n): n \ge 2\} \cup \{\{1\}\}$ really. – Henno Brandsma Jan 16 at 21:34
• Henno Brandsma.Thanks. Understand that {1} is open in [0,1].But you need an open cover in (X,F) , elements of F.Why not just take$\cup [0,1-1/n)\cup [1,2)$? – Peter Szilas Jan 16 at 22:00