Suppose a subset D of a space X meets every nonempty member of a subbasis . Is D necessarily dense in X?

Question

Here is an example that contradicts the question

$X= \{ a, b, c\}$

$S=\left\{ \{a, b, c\}, \{c,b\}, \{c,a\}, \{a,b\}, 0\right\}$

$T=\left\{ \{a, b, c\}, \{a\}, \{b\}, \{a,b\} , \{c\}, \{a,c\}, \{b,c\}, 0\right\}$

Let $D$ be an open set in $X$. $D= \{ c, a\}$.

$D \cap \{c, b\} \cap \{a, b\}=\emptyset$

Since it is equal to the empty set, $S$ must not contain all of its limit points, therefore not closed and not dense.

My question: Is this a good argument to disprove the original question? How can I make this proof more conclusive and formal?

Answer 1

A more extreme example:

A subbase for $[0,1]$ in the usual topology is $\{[0,r), (s,1]: s,r \in (0,1)\}$. Then $\{0,1\}$ intersects all subbasic elements but is very far from dense.

As to your own example: that $D$ works for $S$ is clear. And every dense subset must contain all isolated points (it must intersect all sets $\{x\}$ that are open), so the only dense subset of $T$ (which is discrete) is $X$ itself .And $D \neq X$.

Answer 2

You should define your terms: I.e. "Let $S$ be a sub-base for a topology on $T$ on $X$." Point out that $T$ is the discrete topology on $X.$ That is, every subset of $X$ belongs to $T.$ Now the def'n of "$D$ is dense in $X$ with topology $T$ on $X$" is that $D\subset X$ and that $D\cap t\ne \phi$ whenever $\phi \ne t\in T.$

So in your example, point out that $D=\{a,c\}$ meets every non-empty member of $S,$ but $\{b\}\in T$ and $D\cap \{b\}=\phi.$

This is assuming that $a\ne b\ne c\ne a$, which should be stated at the beginning.

Remark: If $V$ is any collection of subsets of $X$ such that $\cup V=X$ (that is, every member of $X$ belongs to at least one member of $V$) then $V$ is a sub-base for a topology on X. So there are many other counter-examples. E.g. $S=\{(-\infty,x): x\in \mathbb R\}\cup \{(y,\infty):y\in \mathbb R\}$ is a sub-base for the usual topology on $\mathbb R.$ Let $D=\mathbb Z$. Then $D$ meets every member of $S$, but $D$ is not dense in $\mathbb R$ because its complement $\mathbb R$ \ $D$ is open and not empty.