what happens if you use de descrete metric is that any set is open(descrtete toplogy) so any function f:R->R will be continous.(look for definition of continuity in topological spaces)
f:R->R
f(x) is 1 on Q and 0 elsewhere
Show f is not continuous at any point of R.
What about if the usual metric on R is replaced by the discrete metric?
I know that if f(x) takes 1 on Q and 0 elsewhere(Q complement), then by the density property there's infinite number of points between any values of Q, so the graph will keep jumping from 1 to 0 and back, so that's obviously not continuous. Have no idea how to show that in an actual proof though.
Also, I don't really understand replacing a metric by some other metric. What happens id R is replaced by the discrete metric?
I assume that is intially under the usual topology. Then, suppose that is continuous at . Then, since are both dense in there exists sequences in each such that . Thus, . That looks strange, huh?
Also, if is a space with the discrete topology then any mapping is continuous regardless of or . This is clear since if is open then but since every subset of is open we see in particular that is open.
Also, if is a space with the discrete topology then any mapping is continuous regardless of or . This is clear since if is open then but since every subset of is open we see in particular that is open.[/quote]
How do I know that every subset of X is open? Thanks so much.
How do I know that every subset of X is open? Thanks so much.[/QUOTE]
That's what the discrete topology means. The discrete metric gives rise to the discrete topology. To see this merely let have the discrete metric and let be arbitrary. Then, given any we have that contains at most one point of and so ( is the derived set...the set of limit points). Since was arbitrary it follows that . Thus, trivially and thus is closed....BUT! This implies that if then is open since is closed (since any subset of is closed)