Question

let X be a metric space .Show that every subset of X is open ​

Answer 1

Step-by-step explanation:

The discrete metric just says that

d(x,x)=0

d(x,y)=1, x≠y

So say your ball has radius r. If r<1 then the only point it contains is the point it's centred on. So any single point has a ball of some radius around it containing only that point. This is the same thing as B0<r<1(x)={x}, so we know that every singleton is open. And now we're actually done! Since now we know that any point x in a set A has a ball containing it, because we can always construct a ball that only contains x! Since all sets are open, their complements are open as well. This implies that all sets are also closed hope it is useful to you ..☺️