Did I show that this is a basis for a topology on $X=\Bbb N \cup \{-1/n | n \in \Bbb N \}$ correctly?

Question

Suppose $X=\Bbb N \cup \{-1/n | n \in \Bbb N \}$

Given $a,b ∈ X $ above, $b > −1,$

let $U_a := \{x ∈ X |a< x\},$

$U_b := \{x ∈ X |x < b\},$

and for $c,d ∈ X, c < d,$

let $U_d^ c := \{x ∈ X |c < x < d\}.$

Finally, let $B := \{U_a |a ∈ X\} ∪ \{U_b |b ∈ X,b > −1\} ∪ \{U_d^ c |c,d ∈ X,c < d,U_d^ c \neq ∅\}.$

I want to show that B is a basis for a topology on X.

To prove this I am using the definition which states that for a non - empty set X , a basis for a topology on X is a collection of sets $B$ s.t

1) $\forall x\in X, \exists\beta \in B, s.t. x\in \beta$

2)if $x \in \beta_1 \cap \beta_2, \beta_1, \beta_2 \in B$ then $ \beta_3 \in B, s.t. x\in \beta_3 \subset \beta_1 \cap \beta_2 $

Here is my attempt at a solution so far

1) If $x \in X$ then $x\in \{1,2,3,4,...\}\cup \{-1,-1/2,-1/4,....\}$

First consider $x\in \{1,2,3,4...\}$, any such x is also an element of some $U_a$ (or some $U_c^d$)

Next consider $x\in \{-1/2,-1/3,-1/4...\}$ any such is also an element of some $U_a$ (or some $U_c^d$)

finally let $x=-1$ this x is an element of $U^b$

therefore condition one is satisfied

2)suppose $x\in \beta_1 \cap\beta_2 $

Say $\beta_1=U_{a_1}\cup U^{b_1} \cup U_{d_1}^{c_1}$

$\beta_2=U_{a_2}\cup U^{b_2} \cup U_{d_2}^{c_2}$

The intersection of these sets will give

$\beta_1 \cap\beta_2 =U_{max\{a_1,a_2\}}\cup U^{min\{b_1,b_2\}} \cup U_{min\{d_1,d_2\}}^{max\{c_1,c_2\}}$

Suppose $x\in U_{max\{a_1,a_2\}} $

Then if

$\beta_3=U_{a_3}\cup U^{b_3} \cup U_{d_3}^{c_3}$ and $a_3\in (a_1,a_2),b_3\in (b_1,b_2), c_3\in (c_1,c_2), d_3\in (d_1,d_2) $

we will have $x\in \beta_3 \subset\beta_1 \cap\beta_2$

so condition 2 is satisfied.

I have two questions:

1) Does my proof look okay ?

2) What topology does this basis generate ( and if it's not to large a question , how do we find said topology given this basis ?)?

Answer 1

U$_a$ has been defined in two different ways.
Define the lower ray U$_a$ = { x : x < a },
the upper ray U$^a$ = { x : a < x }, and
the interval U$_b^a$ = U$_b$ $\cap$ U$_a.$

The rays are a subbase for a topology.
The rays and the intervals are a base.
The topology is unions of rays and intervals.

This is the linear order topology (LOT).
Every multipoint linear order can be given this topology.

X is a linear order which, with the linear order topology, is homeomorphic to $\omega_0 + \omega_0$ or two copies of N, one right after the other.

Your proof appears correct. It is however notationally awkward. The usual notation for LOT's is the interval notation.