[Math] Are the sets in a basis open in the topology the basis generates

general-topology

A collection $\mathcal{B}$ of subsets of a set $X$ is called a basis if:

  1. For each $x\in X$, $\exists B\in\mathcal{B}$ with $x\in B$.
  2. If $x\in B_1\cap B_2$ for $B_1,B_2\in\mathcal{B}$, then $\exists B_3\in\mathcal{B}$ with $x\in B_3\subset B_1\cap B_2$.

Let $\mathcal{J}$ be the topology on $X$ generated by $\mathcal{B}$:

  • $U \subset X$ is open (that is, $U \in \mathcal{J}$) if $\forall x\in U\; \exists B\in\mathcal{B}:x\in B\text{ and }B\subset U$.

So are basis elements open?

Let $U\in\mathcal{B}$. I need to show that $\forall x\in U$ there exists a basis element that contains $x$ and this basis element must itself be contained in $U$

Best Answer

It has just occurred to me that if we choose $U$ as the basis element then $U\subset U$ and that works to show it is open.

If this were not the case (we demanded a proper subset) then we could always find a smaller basis element, thus we couldn't have a finite basis, so at some point we must accept the relation is not strict.

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