Show that $B=\{[a, b)\subset R \mid a<b\}$ is basis of topology on $R$

Question

Please check my proof

let

$\bigcup_{i=1}^{n}[a_{i},b_{i})$

such that $a\leq U\leqslant b$

We must prove for each $x\in U$ there is open set V in B such that

$x\in V\subset U$

let $x\in U$ and we define

$V=\bigcup_{j=1}^{n} (a_{j},b_{j})$ $ a\leq a_{j}\leq b_{j}\leq b$

since we define $V$ as collection of open interval then it's open set

then for each $x\in U$ there exist s some open subset in $V$ such that $x\in V$

since we define $V=\bigcup_{j=1}^{n} (a_{j},b_{j})$ $ a\leq a_{j}\leq b_{j}\leq b$

the $x\in V\subseteq U$

then $B=\{[a, b)\subset R \mid a<b\} $ is basis of topology on $R$

Answer 1

You must check the two conditions:

a. $\bigcup \mathcal{B} = \mathbb{R}$: Let $x \in \mathbb{R}$. Then $x \in [x,x+1)$ and $[x,x+1) \in \mathcal{B}$, so we're done.

b. For all $B_1, B_2 \in \mathcal{B}$ and all $x \in B_1 \cap B_2$, there exists $B_3 \in \mathcal{B}$ such that $x \in B_3 \subseteq B_1 \cap B_2$: let $B_1 = [a,b)$ and $B_2 = [c,d)$. If $x \in B_1 \cap B_2$, this means that $a \le x < b$ and $c \le x < d$ which is equivalent to $\max(a,c) \le x < \min(b,d)$. So we can take $B_3 = [\max(a,c), \min(b,d)) \in \mathcal{B}$ for all $x \in B_1 \cap B_2$ and we're done.

Your proof makes no sense at all.