Lindelöf and second countable spaces

Question

Can anyone give me some examples and non examples of Lindelöf or second countable space and spaces that is Lindelöf but not second countable? And I understand the definition but find it is hard to visualize and imagine. I have tried google it but it turns out I only found some silly examples like finite set or empty set.

In general, how can one construct a topological space that is Lindelöf or second countable?

Someone in stack exchange said the real line with discrete topology is Lindelöf, but I do not think so. We can simply construct an open cover defined by the collection of all the singleton set. And this open cover is well defined since singleton set is open in discrete topology. Hence, by definition it is not Lindelöf.

Last question, is (0,1) in the real line equipped with usual topology Lindelöf? I think it is Lindelöf but I could not give any formal proof. (0,1) fails to be a compact set since we can construct an open cover defined by (1/n,1-1/n) but this open cover does not work so well for arguing for Lindelöf property since quotient number is dense in (0,1). So intuitively I think it is Lindelöf.

I wrote a pretty long question. My mothertongue is not English. Hopefully, you guys can understand me.

Answer 1

The Sorgenfrey line, also called the lower-limit topology on the real line. It has a basis of intervals $[a,b)$ (or some authors prefer $(a,b]$, upper-limit topology).

It is hereditarily Lindelof (and hereditarily separable, i.e. every subspace has a countable dense subset), but it is not second countable.

Real line with discrete topology is not Lindelof.

$(0,1)$ with the usual topology is Lindelof, it is homeomorphic to the real line. The real line is $\sigma$-compact, that is union of countably many compact subspaces. The real line is the union $\cup_n[-n,n]$, clearly $\sigma$-compact. The interval $(0,1)=\cup_n [\frac1n,1-\frac1n]$ is $\sigma$-compact.

Every $\sigma$-compact space is Lindelof (easy to prove).

In particular, clearly every countable space is Lindelof. But not every countable space is second countable (even with nice separation axioms). For example, the countable sequential fan is not second countable. It is obtained by taking a disjoint family of countably many convergent sequences, and "gluing" their limit points into one limit point, via a quotient map. The result is not first countable at that point. (I couldn't find a suitable online reference to the countable sequential fan, but it has similar properties to the quotient space $\Bbb R/\Bbb N$, which is also not first countable, and likely discussed in most topology books.)

There is an online searchable database (called $\pi$-base), you can make a query asking for Lindelof, not second countable spaces. For many more examples see

https://topology.pi-base.org/spaces?q=lindelof%20%2B%20~Second%20Countable

Every second countable space is Lindelof (you may need to assume some separation axioms, often included in the definitions).

One of the examples at $\pi$-base is the one-point Lidelofication of uncountable discrete space. Take any uncountable set, and a point $p$, and isolate all point but $p$. The neighborhoods of $p$ are co-countable (that is, they have a countable complement). The definition easily implies that the result is a Lindelof space. But it is not first-countable at $p$, and hence not second countable. Often (in the case when "uncountable" is taken to be the first uncountable cardinal) this space is described as the set of all countable ordinals $\omega_1=\{\alpha:\alpha<\omega_1\}$ together with the first uncountable ordinal $\omega_1$, so $X=[0,\omega_1]=\{\alpha:\alpha\le\omega_1\}$, with all countable ordinals $\alpha<\omega_1$ isolated, and with basic neighborhoods of $\omega_1$ of the form $(\alpha,\omega_1]$, with $\alpha<\omega_1$. This is, in addition, an example of a Linearly Ordered Topological Space (LOTS), with only one non-isolated point. (It is a LOTS under a somewhat different order, one may insert a decreasing sequence in front of every limit ordinal.)

The last (and some of the previous examples) are not second countable, because they are not even first countable (and, for the one-point Lindelofication the point $p$ is not even a $G_\delta$ point, that is, it is not the intersection of any countable family of open sets, and the space is not $\sigma$-compact). On the other hand the Sorgenfrey line is perfectly normal: Every closed set (and in particular every point) is the intersection of a countable family of open sets (the proof that the Sorgenfrey line is hereditarily Lindelof uses this). But it is not second countable, since for every basis, and for every $x$ there must be a basic element $B_x$ with $x\in B_x\subseteq[x,\infty)$ and clearly if $x\neq y$ then $B_x\neq B_y$.

An example of a space that is separable but not second countable and not Lindelof is the Moore, or Niemytzki plane (also called tangent-disk space, usually available in topology texts).

There are also compact spaces (which of course is stronger than Lindelof) that are not second countable. One such example closely related to the Sorgenfrey line is the Alexandrov double arrow space, also called split interval. Another is the Alexandroff double circle. Note that every second countable space in hereditarily Lindelof (since every subspace is second countable, and hence Lindelof). The Alexandroff double circle is compact (and hence Lindelof), but has an uncountable discrete subspace, which of course is not Lindelof. Hence the Alexandroff double circle is not second countable.

Answer 2

An example of a Lindelöf non-second countable space, which has some additional nice properties, was constructed/discovered during the Prague 1961 Topological Conference (by wh). The point-set is the unit disc

$$\ B(\mathbf 0\,\ 1)\ := \ \{p\in\mathbb R^2: |p|\le 1\} $$

The neighborhoods of the points $\ p\ $ of the disk, with $\ |p|<1,\ $ are the ordinary Euclidean. In the case of $\ |p|=1,\ $ a base neighborhood, $\ N_{a\,b}(p),\ $ is determined by points $\ a\ b\ $ such that $\ |a|=|b|=1\ $ and $\ a\ne p\ne b\ne a.\ $ This neighborhood consists of points which are between the chord which connects $\ a\ $ to $\ p\ $ and the unit circle, together with a similar one for $\ b\ $ and $\ p\ $ (the arcs $\ ap\ $ and $\ pb\ $ are such $\ a\ $ does not belong to arc $\ pb,\ $ nor $\ b\ $ to $\ ap.$)

Note: Following that Prague conference, my example was published in a paper by A.Archangielski and W.Holsztyński (there is only one paper by these two authors). I've solved a respective question asked by Archangielski.

Answer 3

Lindelöf and second countable are saying that a space is "small" in some sense; so one way to find non-examples is to take products of lots of spaces, such products (or powers) are "big".

$\Bbb R^I$ is not Lindelöf for $I$ uncountable. (it's also not first countable at any point). It is not normal (which is one of the easier ways to see it's not Lindelöf). Of course you're right that the discrete reals are not Lindelöf (take the open cover by singleton sets). BTW, it's a non-trivial fact that if $I$ has size at most that of $\Bbb R$, this product is still separable, so it's also an example of a separable non-Lindelöf space for such $I$.

$[0,1]^I$ for $I$ uncountable is compact (Tychonoff's theorem) so Lindelöf but not first countable at any point too, so certainly not second countable either. But it is normal, of course. And separable iff $|I| \le |\Bbb R|$.

For metric spaces: Lindelöf, second countable and separable are equivalent properties (see a more general fact in my answer here. So $(0,1)$ in the usual topology is certainly Lindelöf, as the rationals in it are dense.

Answer 4

Let's first recall the following properties of second countable spaces.

Theorem: Let X be a second countable space. Then

1. X is a first countable space.
2. X is a separable space
3. X is a Lindelof space

An example of a Lindelof space which is not second countable is the uncountable excluded point space, with the topology defined by:

$$\mathcal{T}_p = \{U \subseteq X : U = X \text{ or } p\notin U \}$$

An explicit example would be $X = (\mathbb{R},\mathcal{T})$. This space is first countable, Lindelof, but not separable and hence not second countable. The reason why $X$ is Lindelof is simply because any open cover $\mathcal{U}$ of $X$ will be written as the union $\mathcal{U} = \mathbb{R} \cup \{ \mathcal{U}_{\alpha} \}_{\alpha \in \mathcal{I}}$, where $\mathcal{I}$ is an index set. The reason why any open cover will include the whole space $\mathbb{R}$ is simply because $\mathbb{R}$ is the only open set containing $p$, and so it must be in the open cover. But, one can simply take the finite subcover $\{ \mathbb{R} \}$, which is obviously countable (has cardinality 1), and obviously covers $\mathbb{R}$. This shows that $X$ is Lindelof.

$X$ is not separable. To see this, note that the closure of any proper subset $V \subset \mathbb{R}$ is simply the concatenation $\bar{V} = V \cup \{p\}$, so the smallest dense subset in this space is $\mathbb{R}\backslash \{p\} $, which is clearly uncountable, but this means that $X$ is not separable, and by the theorem above, it is not second countable, as every second countable space is separable (The converse is not true).