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!

  • $\begingroup$ See the properties of this $\endgroup$ – 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.

  • $\begingroup$ Taemo.A question, please.Would one not have to show that a open (U_a \in F!!) cover ( Union (U_a)) does not have a finite subcover You have {1) in your union, {1} is not an element of F, i.e.not an open set.Thanks, Peter $\endgroup$ – Peter Szilas Jan 16 at 16:49
  • $\begingroup$ @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. $\endgroup$ – Henno Brandsma Jan 16 at 21:34
  • $\begingroup$ 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)$? $\endgroup$ – Peter Szilas Jan 16 at 22:00

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