A Lindelöf space is a topological space in which every open cover has a countable subcover . If $(a,b)$ is an open subset of the reals , then $(a,b)$ is a Lindelöf space . Now , $A={(a,a+k),(a+k-1/n,b)}$ is an open cover for some natural number $n$. There should be a subset of $A$ that still cover $(a,b)$ . I don't know what goes wrong here .

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    $\begingroup$ Please explain your definition of $A$, it is not clear. My guess is that you mean $A = \{ (a+k-1/n,b) : k,n \in \mathbb{N} \}$ or something to this effect. But this is a countable cover, so it is its own subcover. $\endgroup$
    – Ian
    Oct 26 '14 at 20:56
  • $\begingroup$ It really isn’t clear what you’re asking. First, you need to explain $A$ more clearly: is it just two sets, and if so, what are $k$ and $n$, or is it some larger collection? $\endgroup$ Oct 26 '14 at 20:57
  • $\begingroup$ That is itself a countable cover, so surely it is its own countable subcover? $\endgroup$
    – JCW
    Oct 26 '14 at 20:58
  • $\begingroup$ A is a set of subsets of $(a,b)$ that constitute an open cover . I don't know how to write the symbol of set in latex . k is a real number between $a$ and $b$ and n is a large natural number . The point is to obtain an open cover of (a,b) $\endgroup$
    – dcdc
    Oct 26 '14 at 21:01
  • $\begingroup$ @dcdc If you want to prove that $(a,b)$ is a Lindelöf space, you don't get to choose your open cover; you have to start with some arbitrary cover $\mathcal{C}$ and show that you can always pull out a countable subcover. The subcover may look like $A$ as you have defined above, or it may not. Contradiction might be an easier route to prove this. $\endgroup$
    – graydad
    Oct 26 '14 at 21:02

This one is easy: The definition of Lindelof space does not demand a proper open subcover. In particular, as in your troublesome example, if an open cover $C$ of space $S$ is the union of a finite (or indeed a countably infinite)set of open sets, then that some set of open sets can be used to say that $C$ does not demonstrate that $S$ is not Lindelof.

The meatier cases are when you have an uncountable open cover of $S$; then that open cover itself is not admissible and you need to have a countable (thus proper) covering subset of it.


The only case we are interested in is if $\mathcal{C}$ is an uncountable open cover of $(a,b)$ so let $$\mathcal{C} = \{C_\alpha \}_{\alpha \in \lambda}$$ be such an open cover. Now let $x_1 \in (a,b)$. Then there must be some $C_{\beta} \in \mathcal{C}$ such that $x_1 \in C_{\beta}$. For convenience we will define $C_{\beta} =C_{x_{1}}$. Since $C_{x_{1}}$ is an open set, we can find an open interval $(q_1,r_1)$ with $q_1,r_1 \in \Bbb{R}$ such that $x \in(q_1,r_1) \subset C_{x_{1}}$. Now we can continue this process iteratively, letting $$x_{k+1} \in (a,b) \backslash \left(\bigcup_{i=1}^k (q_i, r_i)\right)$$ to form the collection of subsets $\{C_{x_i} \}_{i=1}^{k+1} \subset \mathcal{C}$.

Case 1: If for some $L \in \Bbb{N}$ we have $$(a,b) \backslash \left(\bigcup_{i=1}^L (q_i, r_i)\right) = \emptyset$$ then clearly $$(a,b) \subset \bigcup_{i=1}^L (q_i, r_i)$$ and hence $\{C_{x_i} \}_{i=1}^{L}$ is a finite subcover of $\mathcal{C}$ as $(q_i,r_i) \subset C_{x_i}$ for each $i \in \Bbb{N}$.

Case 2: Not case $1$. It remains to be shown that by continuing the iterative method above you will end up with a countable subcover. This case is a bit trickier. I lean towards contradiction personally, but I will leave the remainder of the proof to you.


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