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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 $\mathbb{R}$.

2.

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.

3.

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.

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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.

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  • $\begingroup$ 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! $\endgroup$ – Chris Culter 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

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