Topology which doesn't contain any singleton set I found the same question here but here answer is given opposite
How can I find a subbasis of a topology which does not contain a singleton set?
Let $(X,T)$ be a topological space. Find a subbasis $S$ for $T$ which does not contain any singleton sets.
$1.$ If $X$ is finite.
$2.$ If $X$ is infinite.
My attempt : For $1).$ take discrete topology$ T$ and $X= \{ a,b,c\}$
Now $ S$ is a subbasis for $(X,T)$  if finite intersection  of member of $S $ give $\{a\}$ , $\{b\}$ and $\{c\}$. so $S=\{ \{a,b\} ,\{b,c\} ,\{a,c\} \}$ which doesnot conatin any singleton sets
Now $ \{a,b\}\cap\{b,c\}= \{b\} $, similarly it will give $\{a\}$ and $\{c\}$
Is its true ?
For $2).$ Im thinking like this take   Usual topology $T$ and $X= \mathbb{R}$
Take subbasis $S=[a,\infty]$  because $[a,x]$ and $[ x, \infty]$ will form subbasis which don't contain any singleton set
Is its true
 A: If $X=\{0,1\}$ with topology $\{\emptyset, \{0\}, X\}$ (the Sierpiński topology), then no such subbase exists. This is easy to see in this case, as there are only 8 subsets of the topology and only those with $\{0\}$ in it (e.g. $\mathcal{S}=\{\{0\}\}$ itself) are subbases for $\mathcal{T}$, or use the argument from the following example..
For an infinite counterexample, take $X=\Bbb N$ and the topology $$\mathcal{T}= \{\emptyset,X\} \cup \{\downarrow m = \{n \in \Bbb N: n \le m\} \mid m \in \Bbb N\}$$ (the down-topology on $\Bbb N$) and show that all subbases for $\mathcal{T}$ include $\{0\}$ (because it is the minimal element by inclusion in the topology, which is linearly ordered). The infinite example can be seen as a generalisation of the Sierpiński example.
A: If $[a,\infty]$ is using the standard notation, it is a closed set.
Then while $[a,\infty] \cap [-\infty,a] = \{a\}$ it is not the case that $\{a\}$ is an open set.
A: Do you mean find a topological space $(X,\cal U)$ so that $\forall x \in X$, $\{x\} \not \in \cal U$?
If so, $\Bbb N$ (the natural numbers) with the cofinite topology is an example, I believe.  (The cofinite topology on $\Bbb N$ consists of all $U\subset \Bbb N$ so that $\overline{U}$ is finite, together with $\emptyset$.  Of course, part of the exercise - I assume this is an exercise - is to show that this defines a topology on $\Bbb N$.)
