Path connectedness of the product topology of Sorgenfrey and cocountable topology Let $\mathcal{T} = \mathcal{T_{[,)}} \times \mathcal{T_{CC}}$ I want to find the connected components of this topology, I already know that Sorgenfrey is not path connected and is totally disconnected. Moreover, the cocountable topology is also not path connected.
Due to this, I cannot directly create a path combining paths of both topologies.
So, is there any path connected component in the product topology?
Thanks in advance.
 A: Any path-connected component must be connected.
Suppose $\langle a,b\rangle$ and $\langle c,d\rangle$ are elements of $\mathcal{T} = \mathcal{T_{[,)}} \times \mathcal{T_{CC}}$.
If $a\neq b$, then $\langle a,b\rangle$ and $\langle c,d\rangle$ are not in the same connected component. Therefore, they are not in the same path-connected component.
But if $a=b$ and if there is a path $\gamma$ from $\langle a,b\rangle$ to $\langle c,d\rangle$,
by projecting to the $\mathcal{T_{CC}}$ part of $\mathcal{T}$ you can see that there is a path from $c$ to $d$ within $\mathcal{T_{CC}}$. This can be the case only when $c=d$.
Therefore, the path-connected components of $\mathcal{T}$ are precisely singletons.
A: The path components of the product are the products of path components.
Since Sorgenfrey is totally disconnected, all you need  is  to find the path components of the cocountable topology.

Proof:
Suppose $X = \bigcup_i X_i$ and $Y = \bigcup_j Y_j$ are the path components. Then we can write $Z = \bigcup_{i,j} X_i \times Y_j = \bigcup_{i,j} Z_{ij}$.
I claim $Z_{ij}$ are the path components of the product.
It is straightforward to see each $Z_{ij}$ is path connected.
Now suppose $z(t) = (x(t),y(t))$ is a path from some point in $Z_{ij}$ to some point in $Z_{kl}$. Then we have
$$x(0) \in X_i \qquad y(0) \in Y_j\qquad  x(1) \in X_k\qquad  y(1) \in Y_l$$
If $Z_{ij} \ne Z_{kl}$ then without loss of generality $i \ne k$. Since the projection is continuous $x(t)$ is a path from $X_i$ to $X_k$ which is not allowed. So no such path $z$ exists.
