# Topological Subspace of Real Number Space is Lindelöf

## Theorem

Let $\left({\R, \tau_d}\right)$ be the real number line considered as a topological space under the usual (Euclidean) topology.

Let $W$ be a non-empty subset of $\R$.

Then $R_W$ is Lindelöf

where $R_W$ denotes the topological subspace of $R$ on $W$.

## Proof

Let $\mathcal C$ be a open cover for $W$.

Define $Q := \left\{{\left({a\,.\,.\,b}\right): a, b \in \Q}\right\}$

Define a mapping $h:Q \to \Q\times\Q$:

$\forall \left({a\,.\,.\,b}\right) \in Q:h \left({\left({a\,.\,.\,b}\right)}\right) = \left({a, b}\right)$

It is easy to see by definition that

$h$ is an injection.
$\left\vert{Q}\right\vert \le \left\vert{\Q \times \Q}\right\vert$

where $\left\vert{Q}\right\vert$ deontes the cardinality of $Q$.

$\Q$ is countably infinite.

By definition of countably infinite:

there exists a bijection $\Q \to \N$

By definitions of set equality and cardinality:

$\left\vert{\Q}\right\vert = \left\vert{\N}\right\vert$
$\left\vert{\Q}\right\vert = \aleph_0$
$\left\vert{\Q \times \Q}\right\vert = \max\left({\aleph_0, \aleph_0}\right) = \aleph_0$
$Q$ is countable.

By definition of cover:

$W \subseteq \bigcup \mathcal C$

By definition of imion:

$\forall x \in W: \exists U \in \mathcal C: x \in U$

By Axiom of Choice define a mapping $f: W \to \mathcal C$:

$\forall x \in W: x \in f \left({x}\right)$

We will prove that

$\forall x \in W: \exists A \in Q: x \in A \land A \cap W \subseteq f \left({x}\right)$

Let $x \in W$.

By definition of open cover:

$f \left({x}\right)$ os open in $R_W$.

By definition of topological subspace:

there exists U a subset of $\R$ such that
$U$ is open in $R$ and $U \cap W = f \left({x}\right)$

By definition of $f$:

$x \in f \left({x}\right)$

By definition of open set in metric space:

$\exists r > 0: B_r\left({x}\right) \subseteq U$
$\left({x - r \,.\,.\, x + r}\right) \subseteq U$
$\exists q \in \Q: x - r < q < x$ and $\exists p \in \Q: x < p < x + r$

By definition of $Q$:

$\left({q \,.\,.\, p}\right) \in Q$

Thus by definition of open real interval:

$x \in \left({q \,.\,.\, p}\right) \subseteq \left({x - r \,.\,.\, x + r}\right)$
$\left({q \,.\,.\, p}\right) \subseteq U$
$\left({q \,.\,.\, p}\right) \cap W \subseteq f \left({x}\right)$

By Axiom of Choice define a mapping $f_1: W \to Q$:

$\forall x \in W: x \in f_1 \left({x}\right) \land f_1 \left({x}\right) \cap W \subseteq f \left({x}\right)$

By definitions of image of set and image of mapping:

$\forall A \in \operatorname{Im} \left({f_1}\right): \exists x \in W: x \in f_1^{-1} \left[ {\left\{{A}\right\} }\right]$

By Axiom of Choice define a mapping $c: \operatorname{Im} \left({f_1}\right) \to W$:

$\forall A \in \operatorname{Im} \left({f_1}\right): c \left({A}\right) \in f_1^{-1} \left[ {\left\{{A}\right\} }\right]$

Define a mapping $g: \operatorname{Im}\left({f_1}\right) \to \mathcal C$:

$g := f \circ c$

Define $\mathcal G = \operatorname{Im} \left({g}\right)$.

Thus $\mathcal G \subseteq \mathcal C$

By definition of image of mapping:

$\operatorname{Im} \left({f_1}\right) \subseteq Q$
$\operatorname{Im} \left({f_1}\right)$ is countable.
$\left\vert{\operatorname{Im} \left({g}\right)}\right\vert \le \left\vert{\operatorname{Im} \left({f_1}\right)}\right\vert$
$\mathcal G$ is countable.

It remains to prove that

$\mathcal G$ is a cover for $W$.

Let $x \in W$.

By definition of $f_1$:

$x \in f_1 \left({x}\right) \land f_1 \left({x}\right) \cap W \subseteq f \left({x}\right)$

By definition of image of mapping:

$f_1 \left({x}\right) \in \operatorname{Im} \left({f_1}\right)$

Then by definition of $c$:

$y := c \left({f_1 \left({x}\right)}\right) \in f_1^{-1} \left[ {\left\{ {f_1 \left({x}\right)}\right\} }\right]$

By definition of $f_1$:

$y \in f_1 \left({y}\right) \land f_1 \left({y}\right) \cap W \subseteq f \left({y}\right)$

By definition of image of set:

$f_1 \left({y}\right) = f_1 \left({x}\right)$

Then by definitions of subset and intersection:

$x \in f \left({y}\right)$

By definition of composition of mappings:

$f \left({y}\right) = g \left({f_1 \left({x}\right)}\right)$

By definition of image of mapping:

$g \left({f_1 \left({x}\right)}\right) \in \mathcal G$

Thus by definition of union:

$x \in \bigcup \mathcal G$

Thus the result.

$\blacksquare$