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I'm beginning to study topology and I have troubles solving these exercises (my book has no answers, unfortunately). Could you help me?

Consider lower limit topology $ \mathcal{T}$ generated by basis $ \mathcal{B} = \left\{ [a,b) \ | \ a,b \in \mathbb{R}, \ a<b\right\} $.

I can already prove that this topological space is Hausdorff space, meaning that for any $x,y \in \mathbb{R}, \ x \neq y$ we can find disjoint neighbourhoods, and that $ \{ x \}$ is closed in this topology.

What I can't prove is that:

1) $ \forall x \in \mathbb{R} \ \forall A \subset \mathbb{R} $ closed in $(\mathbb{R}, \mathcal{T} ), \ x \not \in A \ \ \ \exists U, V \in \mathcal{T} \ : \ x \in U, \ A \subset V, \ U \cap V = \emptyset $ (I hope it's comprehensible)

2) Prove that each point in $(\mathbb{R}$ has a countable neighbourhood basis in $(\mathbb{R}, \mathcal{T})$

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Note that this question shows that the lower limit topology is Hausdorff, and this question shows that it is normal, which would imply the regularity of the space. –  Arthur Fischer Mar 2 '13 at 15:20

1 Answer 1

HINTS:

(1) Let $W=\Bbb R\setminus A$; then $W$ is an open nbhd of $x$, so $[x,y)\subseteq W$ for some $y\in\Bbb R$. Now use the fact that $[x,y)$ is clopen (= closed and open).

(2) For $x\in\Bbb R$ consider the basic open nbhds $\left[x,x+\frac1n\right)$ with $n\in\Bbb Z^+$. Added: An even better idea is to let $\mathscr{B}(x)=\left\{[x,q):x<q\in\Bbb Q\right\}$; this is still countable, and it makes verification even easier.

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I'm not sure if it's ok, but could we take $x<y<z \in \mathbb{R}$, $A=[y, z)$, then as $U$ choose $[x,y)$ and $V=[y, z+1)$? –  Hagrid Mar 2 '13 at 14:17
    
@Hagrid: You don’t get to choose $A$: all you know about it is that it’s a closed set that does not contain $x$. Let me add a little to the first hint: take $U=[x,y)$, and note that $U$ is not only open, but also closed (why?). What does that tell you about $\Bbb R\setminus U$? And how are $A$ and $\Bbb R\setminus U$ related? –  Brian M. Scott Mar 2 '13 at 14:21
    
$[x, y)$ is clopen because it is a union of elements of the basis and the complement of $[x, y)$, too. This means that $\mathbb{R} \setminus U$ is also clopen. $A$ is a closed set that does not contain $x$, so it must be a complement of an open set containing $x$, so this means that $A \subset \mathbb{R} \setminus U$ –  Hagrid Mar 2 '13 at 14:37
    
@Hagrid: Right. So you can simply let $V=\Bbb R\setminus U$: then $x\in U$, $A\subseteq V$, $U$ and $V$ are open, and $U\cap V=\varnothing$. –  Brian M. Scott Mar 2 '13 at 14:38
    
As for (2), I need to check that $\left[x,x+\frac1n\right)$, $n \in \mathbb{N} ^+$ is a neighbourhood basis $\mathcal{B} (x)$ . There are 3 conditions: (i) $\forall x\in \mathbb{R} : \ \mathcal{B} (x) \neq \emptyset$ and for any $U \in \mathcal{B} (x)$. (ii) If $x \in U \in \mathcal{B} (y)$, then there exists $V \in \mathcal{B} (x)$ such that $V \subset U$, (iii) $\forall U_1, U_2 \in \mathcal{B} (x) \ \exists U \in \mathcal{B} (x) : \ U \subset U_1 \cap U_2$. The first one looks easy. Could you help me with (ii), (iii) ? Or maybe there is another way to prove it's a nbhd basis. –  Hagrid Mar 2 '13 at 15:01

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