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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 $U$

Taumatawhakatangihangakoauauot
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Alec Teal
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  • @ThomasAndrews yes, that would generate the indiscreet topology. – Alec Teal Jan 01 '15 at 12:59
  • Right, I misread what you were defining. – Thomas Andrews Jan 01 '15 at 12:59
  • @ThomasAndrews your comment was a good example though! I wish you hadn't have deleted it (I thought (okay not for long - I am proud to say) about it and concluded it generated the familiar indiscreet topology) – Alec Teal Jan 01 '15 at 13:01
  • It was a pointless comment, and I don't like leaving litter on this site. I thought you were defining a "basic for a topology $\tau$," not a general basis which can be used to define a topology. A subtle distinction. – Thomas Andrews Jan 01 '15 at 13:05

2 Answers2

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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.

Alec Teal
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    Yes, that is exactly it. For each $x\in U$, take $B=U$. Better to use $B\subseteq U$ in the definition above, because it is clearer - some people read $\subset$ as $\subsetneq$. – Thomas Andrews Jan 01 '15 at 13:01
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Yes, the basis elements are open.

A set $X$ and a collection $\mathfrak B$ of subsets of $X$, which satisfy the given properties. Only then do we define the topology $\mathscr T$ using this basis, so until this is done it doesn't make sense to talk about the basis elements of $\mathfrak B$ as being "open".

Now, once we do define our topology, recall that we define a subset $U$ of $X$ to be open if given $x\in U$ there exists $B\in\frak B$ such that $x\in B\subseteq U$. Since this certainly holds when $U=B\in\frak B$, we see that all elements of $\frak B$ are open.

Now starting with some collection of subsets $\frak B$ that generate a topology, calling this topology $\scr T$. So by what we said above, all basis elements are open in $\scr T$ and there should be no issue.

Sandip Agarwal
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