Find 3 topologies on thefive point set X = {a,b,c,d,e} such that the first is finer then the second and the second is finer then the third, without using either the trivial or discrete topology.
Hello,
A topology A is said to be finer than a topology B, if any open set of B is an open set of A.
Knowing that, it's quite easy
Consider the less fine topology, let's say {0,{a},X} (0 designs the empty set)
Then you have to find a topology that contains {a} (it'll always contain 0 and X, because it's the definition) and more sets.
So you can add {b} and {a,b}. You have to add {a,b} because the union of two elements of a topology is in the topology and {a} U {b} = {a,b}
Hence {0,{a},{b},{a,b},X} can be the second finest topology.
You need to find a third one, that contains {a},{b},{a,b}
Let's add {c}
Since it has to be stable by union (for intersection, it's trivial because it's the empty set), you have to add {a} U {c}={a,c}, {b} U {c}={b,c} and {a,b} U {c}={a,b,c}
So the finest topology can be {0,{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c},X}