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Let $\mathbb{J} :=\{1/n: 0< n\in \mathbb{Z}\}$
Let $T_{ir}$ be topology of $\mathbb{R}$ generated by $$\{(a,b)\subset \mathbb{R}:a<b\}\cup\{(a,b) \setminus \mathbb{J}\subset \mathbb{R}:a<b\}$$ (a)Discuss the compactness of a closed interval $[a, b]$ in $(\mathbb{R}, T_{ir})$.
(b)Show that $(\mathbb{R}, T_{ir})$ is connected but not path connected.

For (a), I know that if $0<a$ or $b\le 0$, then $[a, b]$ is compact. How about the case: $a=0<b$?
For (b), I have no any idea about connectedness. I want to show it is not path connected by considering path joining $0$ and $1$ but I want give a rigorous proof.
Please give me all detail. Thank you.

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$\frac{1}{2}(\text{Hint}+\text{answer})$: For compactness, let's consider for instance $[0,1]$. The only thing that can go wrong for compactness is if we try to include our new open subsets in our covering, so let's do that and see what happens. To get almost all of $[0,1]$ we could start by including $I = (-1,2) \setminus \mathbb{J}$. Of course, to include the points $[0,1] \cap \mathbb{J}$, we need to use some of our regular open sets. We don't want them to be so big that it's easy to find a finite subcover, so let's use open intervals $I_n$ that include the point $1/n$ in $\mathbb{J}$ but no other points from $\mathbb{J}$ (you can construct such intervals explicitly if you like). So now, by construction $I \cup \bigcup_n I_n$ will be a covering of $[0,1]$. Does it have a finite subcover?

For connectedness, what have you tried?

For non-path-connectedness, you're on the right track. The only way such an argument would not instead show path-connectedness would be if your path was not continuous, so that's what you want to show. So, what do preimages of open subsets look like? In particular, what about preimages of the new open subsets you've added?

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  • $\begingroup$ For connectedness, I have tried: Let $U$,$V$ $\in T_{ir}\setminus \{\emptyset,\mathbb{R} \}$ s.t. $U \cup V=\mathbb{R}$. I want to show $U \cap V \neq \emptyset$. <br/>Suppose not. If $U$,$V$ $\in T_{std}$, done. WLOG, assume $U\in T_{ir}\setminus T_{std}$. If $ 0\notin U$, then $U\in T_{std}$. So, $0 \notin V$ and hence $V\in T_{std}$. There exist $n$ and $\epsilon$ s.t. $(1/n-\epsilon,1/n+\epsilon)\setminus\{1/n\}\subset U $but $1/n\in V$.(otherewise $\mathbb{J}\subset V$ and $U\in T_{std}$) $\endgroup$ – user134927 Apr 30 '14 at 12:43
  • $\begingroup$ But $V\in T_{std}\implies $there exist open interval $I$ s.t.$\emptyset \neq I\cap (1/n-\epsilon,1/n+\epsilon) \subset U\cap V$.Contradiction.But I am not sure if this answer is correct $\endgroup$ – user134927 Apr 30 '14 at 12:44

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