I'd like to have a check about this exercise, which asks me if $\mathbb{R}$, with three different topologies, is compact

  1. $\tau=${$U \subseteq \mathbb{R}: [-1,1] \subset U$} $\cup$ {$\emptyset$}
    Say if $\mathbb{R}$ is compact.

Let $\mathcal{R}$ be an open cover of $\mathbb{R}$. $\mathcal{R}=\cup_{i\in I}{A_i}$. If $A_i=(-1-i,1+i)$, $i>0$, then this is a open cover, but I can't find a finite subcover because this wouldn't cover the whole $\mathbb{R}$.


Let $\mathcal{B}=${$(a,b): a<0, b>1, a,b \in \mathbb{R}$} the basis which generates the topology $\tau$. Say if $\mathbb{R}$ is compact.

Also here, if I take an open cover $\mathcal{R}=\cup_{i \in I}A_i$, with $A_i=(a-i,b+i)$, $i>0$, their union covers the whole $\mathbb{R}$, but just if it's finite, so the set it's not compact.


Let $\mathcal{B}=${$[a,+\infty): a \in \mathbb{R}$} the basis which generates the topology $\tau$. Say if $\mathbb{R}$ is compact.

I consider the open cover $\mathcal{R}=\cup_{x\leq a}[a,+\infty)$. It's open because union of open sets. If there would be a finite subcover, then I could stop to an $\overline{a}$, such that $[\overline{a},+\infty)$: but in this case this wouldn't cover $(-\infty,\overline{a})$, so I can't find a finite subcover, and $\mathbb{R}$ is not compact.


The general ideas are right, but in (2) and (3), you reuse variables that are not defined in the relevant scope. Specifically, when you say this:

with $A_i=(a-i,b+i)$

the open cover $\mathcal{R}=\cup_{x\leq a}[a,+\infty)$.

That doesn't make sense, because $a$ and $b$ are not fixed numbers that are given to you; they are bound variables used in the definition of the topology.

Bottom line: Don't use the letters "$a$" or "$b$" when you define your $\mathcal{R}$. Ideally, don't use them anywhere. Instead, replace them with constants like $0$, $1$, or $-1$, or find a way to eliminate them altogether.

  • $\begingroup$ So I should write $A_i=(1-i,1+i)$ instead of $(a-i,b+i)$ for (2) and in (3) $\mathcal{R}=\cup_{x\leq 1}[a,+\infty)$, for example? $\endgroup$
    – VoB
    Aug 24 '17 at 9:04
  • $\begingroup$ Something like that! I think you actually want $A_i=(-i,1+i)$ (because the lower limit must be $<0$) and $\mathcal{R}=\cup_{x\leq 1}[x,+\infty)$ (using $x$ instead of $a$). In fact, if you want to be really terse, you can just take $\mathcal R=\mathcal B$ for both problems! $\endgroup$ Aug 24 '17 at 9:09
  • $\begingroup$ Oh yes, sorry ;) Of course i meant $(-i,i+1)$... the same for $\mathcal{R}=\cup_{x\leq 1}[x,+\infty)$ Thanks so much ;) $\endgroup$
    – VoB
    Aug 24 '17 at 9:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.