Help me visualize a Lindelöf space and a corresponding theorem

Question

I am self studying Metric Spaces from the book by Satish Shirali and H.L. Vasudeva. I am having trouble in visualizing how a Lindelöf space looks like. The book defines it as:

A metric space is said to be Lindelöf if each open covering of $X$ contains a countable subcovering.

I am actually having difficulty in wrapping my head around the concept of a countable subcovering inside a covering. A relevant diagram regarding the same would be highly useful. Also, if possible , please give an example of a Lindelöf space and one of a non-Lindelöf space. It would be extremely appreciated.

Besides, there is a proposition regarding Lindelöf spaces, which seems to be all the more abstract.

Proposition: Let $(X, d)$ be a metric space. If $X$ satisfies the second axiom of countability, then every open covering $\{U_a\}_{a \in \Lambda}$ of $X$ contains a countable subcovering. In other words, a second countable metric space is Lindelöf.

Proof: Let $\{G_i : i = 1, 2, \cdots\}$ be a countable base of open sets for $X$. Since each $U_a$ is a union of sets $G_i$ , it follows that a subfamily $\{G_{i_j} : j = 1, 2, \cdots \}$ of the base $\{G_i : i = 1, 2, \cdots \}$ is a covering of $X$. Choose $U_{i_j} \supseteq G_{i_j}$ for each j. Then $\{U_{i_j} : j = 1, 2, \cdots \}$ is the required countable subcovering.

Please help me understand this diagrammatically! I am completely lost.

Thank You !

Answer 1

Ruy already said a lot about motivation that I won't repeat here, but anyway, having a countable subcover (instead of a finite one, as in the definition of compact) for every open cover (which could be huge!) is the "next best thing", it allows us to work with countable covers for which we can do recursion arguments, make sequences etc. And analysis prefers countable things (sequences, series, sigma-finite measures) because most tools are built for that.

A typical non-Lindelöf metric space is $\Bbb R$ in the discrete metric $d(x,y)=1$ when $x \neq y$, then $B(x,1)= \{x\}$ for all $x$ and the open cover $\{x\}, x \in \Bbb R$ of the space has no proper subcover at all: if we omit some $\{p\}$ then $p$ won't be covered anymore, and as $\Bbb R$ is uncountable, this open cover has no countable subcover.

For metric spaces we have that being Lindelöf is equivalent to being separable (having a countable dense subset) and to being second countable. So most of the favourite spaces in analysis are indeed Lindelöf.

The theorem's proof is somewhat chaotic and badly written. Suppose $X$ is a second countable space, with countable base $\mathcal{B}= \{B_n\mid n \in \Bbb N\}$. Let $\mathcal{U}$ be any open cover of $X$. For each $x \in X$ there is some $U_x \in \mathcal{U}$ with $x \in U_x$. This uses that we have a cover. As $U_x$ is open and $x \in U_x$ there is some $n(x) \in \Bbb N$ such that $x \in B_{n(x)} \subseteq U_x$. This uses that $\mathcal{B}$ is a base.

Now let $M=\{n(x)\mid x \in X\}$ which is a countable set as a subset of $\Bbb N$. For each $m \in M$ we pick $x_m$ so that $n(x_m)= m$ and then we note that

$\{U_{x_m}\mid m \in M\}$ is a countable subcover of $\mathcal{U}$. That it's a subset of $\mathcal{U}$ is clear by definition, so let $p \in X$ be arbitrary. Then we have $p \in B_{n(p)} \subseteq U_p$ "by construction", and $n(p) \in M$ by definition. So $n(p) =: m_0 = n(x_{m_0})$ also by construction but then $p \in B_{n(p)} = B_{n(x_{m_0})} \subseteq U_{x_{m_0}}$, so that $p$ is indeed covered.

Bit technical index juggling with some have use of AC (axiom of choice) but the idea is clear: we just use countably many base elements in a refinement of the cover and we use some expansion of that refinement as subcover. Your proof had sort of the same idea, but in other notation.

Answer 2

The objects studied in Math often come in many sizes, some of which might be too big for us to deal with. So we end up introducing notions of smallness in order to rule out the scary monsters we're afraid of.

In Linear Algebra, this is done e.g. by introducing finite dimensionality. In Group Theory, finite groups are often easier to study, and perhaps also the finitely generated ones.

In measure theory, people sometimes prefer to deal with measure spaces whose total measure is finite and, if that is to restrictive, $\sigma $-finite measure spaces are considered instead.

In topology, the most restrictive size limitation is perhaps to stick to finite spaces, but that should probably be thought of as size XXS (extra-extra-small), and hence does not fit too many people!

The next size up the chart is perhaps occupied by compact spaces, which indeed present many of the features of finite sets.

Moving up a notch, one may look at the $\sigma $-compact spaces, namely those which can be expressed as a countable union of compact subsets, aka the $\sigma $-compact spaces.

If $X$ is a $\sigma $-compact space, say $$ X=\bigcup_{n\in {\bf N}}K_n, $$ where the $K_n$ are compact subsets, then every open covering $\{U_i\}_{i\in I}$ of $X$ is also an open covering for each $K_n$, so we may find finite subsets $I_n\subseteq I$, such that the finite collection $\{U_i\}_{i\in I_n}$ covers $K_n$. Clearly $J=\bigcup_{n\in N}I_n$ is countable and $\{U_i\}_{i\in J}$ is a countable subcovering for $X$.

In other words, every $\sigma $-compact space is Lindelöf, so the latter may be thought of as the next size up the size chart for topological spaces.