Coarsest topology
I am reading a book Topology and Geometry by Bredon. I can't understand definition of coarsest topology. Who can explain it to me (with examples)? Thank you!
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$\begingroup$The coarsest topology is the smallest, in the sense of inclusion. Every other topology $T'$ satisfying the condition contains $T$, i.e. $T \subset T'$. Intuitively, bigger/stronger/finer topologies give you more open sets.
A simple example is the trivial topology, which is the coarsest topology possible.
Edit: As a simple exercise, what is the finest topology possible?
$\endgroup$ $\begingroup$Given a set $X$ you can endow it with any topology you'd like let's call it $T_1$, a way to know how continuous function behave on this topology is to see if the open sets of this new topology are also open on some topology you know, say $T_2$.
Definition: If $T_1\subset T_2$ then we say that $T_1$ is coarser (or smaller) than $T_2$, similarlly if $T_1\supset T_2$ we say that it is finer (or larger).
An easy example as user @J.W.Tanner suggest is to consider the indiscrete topology $I(X)$ on a set, then $I(X)\subset T_2$ for every $T_2$ since by definiton a topology must contain $\emptyset$ and $X$.
Let me illustrate how this definition helps us by giving a proposition
Proposition: Let $(X,T_1)$ and $(X,T_2)$ be topological spaces such that $T_1\subset T_2$ then
- The identity map $id_X:(X,T_1)\to(X,T_2)$ is continuous.
- The identity map $id_X:(X,T_2)\to(X,T_1)$ is open
The proof of this is straightforward from the definitions and you should try to prove it yourself.
From this propositon we can say that if you have a continuous map $f:X\to Y$ and we change the topology on $X$ to be finer then the map is still continuous since $id_X\circ f=f$ is a composition of continuous maps.
$\endgroup$ $\begingroup$Compare the trivial topologies of a non-empty set $X$, one being the topology $T$ made up by the open sets $\{ \emptyset, X\}$ of $X$ and the other $T'$ made up by: any subset of $X$ is an open set, and the empty set is also open.
You see that any open set on the $T$ topology (which are only two) is also open on the $T'$ topology, but the reverse is not necessarily true. It's in this sense that the $T$ topology is coarser: I like to think that you can't refine it down around points of the set. It's also called smaller due to giving you fewer tools in the same sense as before.
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