Topological Subspace of Real Number Line is Lindelöf

Theorem

Let $\struct {\R, \tau_d}$ be the real number line with 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 $\CC$ be a open cover for $W$.

Define $Q := \set {\openint a b: a, b \in \Q}$

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

$\forall \openint a b \in Q: \map h {\openint a b} = \tuple {a, b}$

It is easy to see by definition that

$h$ is an injection.

By Injection iff Cardinal Inequality:

$\card Q \le \card {\Q \times \Q}$

where $\card Q$ deontes the cardinality of $Q$.

By Rational Numbers are Countably Infinite:

$\Q$ is countably infinite.

By definition of countably infinite:

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

By definitions of set equality and cardinality:

$\card \Q = \card \N$

By Aleph Zero equals Cardinality of Naturals:

$\card \Q = \aleph_0$

By Cardinal Product Equal to Maximum:

$\card {\Q \times \Q} = \max \set {\aleph_0, \aleph_0} = \aleph_0$

By Countable iff Cardinality not greater than Aleph Zero:

$Q$ is countable.

By definition of cover:

$W \subseteq \bigcup \CC$

By definition of imion:

$\forall x \in W: \exists U \in \CC: x \in U$

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

$\forall x \in W: x \in \map f x$

We will prove that

$\forall x \in W: \exists A \in Q: x \in A \land A \cap W \subseteq \map f x$

Let $x \in W$.

By definition of open cover:

$\map f x$ is open in $R_W$.

By definition of topological subspace:

there exists U a subset of $\R$ such that

$U$ is open in $R$ and $U \cap W = \map f x$

By definition of $f$:

$x \in \map f x$

By definition of open set in metric space:

$\exists r > 0: \map {B_r} x \subseteq U$

$\openint {x - r} {x + r} \subseteq U$

By Between two Real Numbers exists Rational Number:

$\exists q \in \Q: x - r < q < x$ and $\exists p \in \Q: x < p < x + r$

By definition of $Q$:

$\openint q p \in Q$

Thus by definition of open real interval:

$x \in \openint q p \subseteq \openint {x - r} {x + r}$

By Subset Relation is Transitive:

$\openint q p \subseteq U$

Thus by Set Intersection Preserves Subsets/Corollary:

$\openint q p \cap W \subseteq \map f x$

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

$\forall x \in W: x \in \map {f_1} x \land \map {f_1} x \cap W \subseteq \map f x$

By definitions of image of set and image of mapping:

$\forall A \in \Img {f_1}: \exists x \in W: x \in f_1^{-1} \sqbrk {\set A}$

By Axiom of Choice define a mapping $c: \Img {f_1} \to W$:

$\forall A \in \Img {f_1}: \map c A \in f_1^{-1} \sqbrk {\set A}$

Define a mapping $g: \Img {f_1} \to \CC$:

$g := f \circ c$

Define $\GG = \Img g$.

Thus $\GG \subseteq \CC$

By definition of image of mapping:

$\Img {f_1} \subseteq Q$

By Subset of Countable Set is Countable;

$\Img {f_1}$ is countable.

By Surjection iff Cardinal Inequality:

$\size {\Img g} \le \size {\Img {f_1} }$

Thus by Countable iff Cardinality not greater than Aleph Zero:

$\GG$ is countable.

It remains to prove that

$\GG$ is a cover for $W$.

Let $x \in W$.

By definition of $f_1$:

$ x \in \map {f_1} x \land \map {f_1} x \cap W \subseteq \map f x$

By definition of image of mapping:

$\map {f_1} x \in \Img {f_1}$

Then by definition of $c$:

$y := \map c {\map {f_1} x} \in f_1^{-1} \sqbrk {\set {\map {f_1} x} }$

By definition of $f_1$:

$y \in \map {f_1} y \land \map {f_1} y \cap W \subseteq \map f y$

By definition of image of set:

$\map {f_1} y = \map {f_1} x$

Then by definitions of subset and intersection:

$x \in \map f y$

By definition of composition of mappings:

$\map f y = \map g {\map {f_1} x}$

By definition of image of mapping:

$\map g {\map {f_1} x} \in \GG$

Thus by definition of union:

$x \in \bigcup \GG$

Thus the result.

$\blacksquare$

Sources

• 1989: Ryszard Engelking: General Topology (revised and completed ed.)

• Mizar article TOPGEN_6:Lm3