2

I have a lemma for a basis of a topology:

Let $X$ a non-empty set. $\mathcal B\subset \mathcal P (X)$ is basis of any topology on $X$ if and only if:

  1. $X=\bigcup_{B\in \mathcal{B}}B$

  2. $\forall B_1, B_2\in\mathcal B, \ \forall x\in B_1\cap B_2 \ \exists B_3 \mid x\in B_3\subset B_1\cap B_2$.

What if for example $X=\{1,2,3\}$, $\mathcal{B}=\{\{1\},\{2\},\{3\}\}$. What if we have $x\in \{1\} \cap \{2\} = \emptyset$. Then how do we find a $B_3$?

amWhy
  • 209,954
  • 3
    what is we have $x\in\emptyset$? That never happens, so you don't need to worry about it. Axiom 2 says nothing unless $x\in B_1\cap B_2$, which implies that $B_1\cap B_2\ne\emptyset$. – David C. Ullrich Jul 10 '21 at 11:15
  • @DavidC.Ullrich Okay, I think I've been confused with the empty set. Also by the definition of a basis, a topology $\mathcal{T}$ has basis $\mathcal{B}$ if its open sets are the unions of set from $\mathcal{B}\subset \mathcal{T}$. Why doesn't the basis have an empty set? –  Jul 10 '21 at 11:19
  • 3
    @use80085 The empty union $\bigcup_{B \in \emptyset} B$ is equal to the empty set $\emptyset$. See, for example, this. – Amit Rajaraman Jul 10 '21 at 11:29

1 Answers1

2

$B_1 \cap B_2 = \emptyset$ so any statement that begins $\forall x \in B_1 \cap B_2$ is automatically true, as there are no points to worry about, see here, e.g.

Henno Brandsma
  • 242,131