Does it make sense to talk about sub-topologies?

Question

Upon examining the real line with the finite complement topology $\tau_{c}$, an open set $O$ is a union of certain open intervals in $\mathbb{R}$. Since $\mathbb{R}$ itself is the single interval $(-\infty,\infty)$, the empty set is the empty union of intervals, and any other set is of the form $O=(-\infty,p_1)\cup (p_1,p_2)\cup\dots\cup (p_{n-1},p_n)\cup (p_n,\infty)$ where $p_1<p_2<\dots <p_n$. Well, the usual topology on the real line $\tau$, is made up of all union of open intervals. So it seems like (and forgive me if this isn't the proper notation) that we could say $\tau_c\subset\tau$. I can't seem to find anything about "sub topologies" online though. Does this idea make sense/ is it any useful? Is there a name for this situation? This is my first semester in topology, so I'm not sure what to look up to find anything about it (but I've tried!).

Answer 1

The 'sub-topologies' that you mention are known as coarser topologies.

As an example you may also take the Sorgenfrey line $\mathbb{R}_l$ and $\mathbb{R}$ with the usual topology. Every open set in $\mathbb{R}$ is open in $\mathbb{R}_l$ too, however there are open sets in $\mathbb{R}_l$ which are not open in $\mathbb{R}$. So $\mathbb{R}$ is coarser than $\mathbb{R}_l$.

Answer 2

$\tau_c$ is not a topology. For example $(-\infty,0)\cup \bigcup_n (\frac 1 {2n}, \frac 1 {2n-1})\cup (0,\infty)$ does not belong to it though $(-\infty,0)\cup (\frac 1 {2n}, \frac 1 {2n-1})\cup (0,\infty)$ does not belong to it for each $n$.