Conectedness in the topology of the included subset Considering the following family of subsets
$$T=\{U:A\subseteq U\}\cup\{\varnothing\}. $$
I have easily proved it's a topology.
How can I find its connected components and which subsets are conected?
 A: This topology is called $A$-include topology. Notice that for any open sets $U_1$ and $U_2$, $U_1\cap U_2\supseteq A$, so if $A\ne \varnothing$, then the topology is connected and there is only one component.
Also note that if $B\subset X$ is not connected, then $B=(B\cap U_1)\cup (B\cap U_2)$ with $B\cap U_1\cap U_2=\varnothing$ which shows that $B\cap A=\varnothing$. So any subset $B$ with $B\cap A=\varnothing$ is disconnected.
If $A=\varnothing$, then the topology is just discrete topology and hence totally disconnected.
A: Denote the topology you defined by $\mathcal{T}(A)$ and note that it's also a topology (the discrete one) if $A=\emptyset$, but that $\mathcal{T}(A)$ is connected iff $A \neq \emptyset$, because then all non-empty open sets contain $A$ and so there are no disjoint non-empty sets at all, so no disconnections of $X$.
When $B$ is a subspace of $X$, it's clear that the subspace topology on $B$ induced by $\mathcal{T}(A)$ is exactly $\mathcal{T}(A\cap B)$ on $B$, so $B$ is a connected subspace iff $A \cap B \neq \emptyset$ except possibly when $B$ is empty or a singleton, and then $B$ always connected, regardless of the topology on $X$.
So Qurultay's claim that all subspaces of $X$ are connected is false: e.g. $X=\Bbb N$ with $A=\{0,1,2\}$ has $B=\{3,4,5,\ldots\}$ as a discrete and thus disconnected subspace.
