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There is this following situation:

For $\mathbb{R}$ we consider the family $\mathcal{S}$ of subsets consisting of all the intervals of type $(m,M)$ with $m<M<0$, all the intervals of type $(m,M)$ with $0<m<M$ (both $m,M\in\mathbb{R}$) and the interval $[-1,1)$. We denote by $\mathcal{T}$ the smallest topology on $\mathbb{R}$ containing $\mathcal{S}$.

Now I have to find an interval of type $[a,b]$ with the property that, together with the topology induced from $(\mathbb{R},\mathcal{T})$, is not compact.

I also have to find an interval of type $(a,b)$ with the property that, together with the topology induced from $(\mathbb{R},\mathcal{T})$, is not connected.

I already found the basis for this topology, which is $$ \mathcal{B}=\mathcal{S} \cup \{[-1,M): M\in \mathbb{R}, -1<M<0\}.$$

Can somebody help me? Thanks a lot!

jbuser430
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1 Answers1

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For an open interval that cannot be compact under $(\mathbb{R},\mathcal{T})$, look at $U = [a,-1]$ with $m < a < -2$. Then define $A_{n} = (m, -1 - \frac{1}{n}), n \in \mathbb{Z}^{+}$. Then $\bigcup_{n}A_{n} \cup [-1, 1)$ cover $U$, but contain no finite subcover.

For such an interval that cannot be connected, look at $V = (a, 1)$. Simply take $a = m, M = -1$ for $m<M<0$. Then $(m, -1) \cup [-1,1)$ constitute an open separation of $V$.

Alex
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