19-01-2023

Problem: Let $\mathcal{T} _1$ and $\mathcal{T} _2$ be two topologies on a space $X$ with $\mathcal{T} _1\subset \mathcal{T} _2$. If $\left( X,\mathcal{T} _1 \right) $ is connected, then comment on the connectedness of $\left( X,\mathcal{T} _2 \right) $.

Solution: If $X$ has at least two points, then it need not be connected with respect to the $\mathcal{T} _2$ topology. For example, \[ \mathcal{T} _1= \text{Indiscrete topology},~ \mathcal{T} _2 = \text{discrete topology}. \] It is clear that $\mathcal{T} _1\subset \mathcal{T} _2$, as the discrete topology is the finest topology. Since $X$ have more than one points, it is not connected, as if $x_1,x_2\in X$, then $\{x_1\}$ and $\{x_2\}$ both are closed and open.


The other way around is true, that is if $\left( X,\mathcal{T} _2 \right) $ is connected, then $\left( X,\mathcal{T} _1 \right) $ is also connected. For a proof of this fact, let us suppose that $\left( X,\mathcal{T}_1 \right) $ is not connected. This implies, there exists $A,B\in \mathcal{T} _1$ such that both are non-empty, proper subsets of $X$ and \[ X = A\cup B. \] As $\mathcal{T} _1\subset \mathcal{T} _2$, so $A,B\in \mathcal{T} _2$, this implies $A$ and $B$ gives a separation for $\left( X,\mathcal{T} _2 \right)$ also, which implies $\left( X,\mathcal{T} _2 \right) $ is also disconnected, a contradiction. Therefore, $\left( X,\mathcal{T} _1 \right) $ is connected.