- #1

Ted123

- 446

- 0

[PLAIN]http://img855.imageshack.us/img855/5949/metric.jpg

If [itex]X,Y[/itex] are sets and [itex]f:X\to Y[/itex] is a function with [itex]B\subset Y[/itex], the

If [itex]d_X, d_Y[/itex] are metrics on [itex]X,Y[/itex], continuity of [itex]f[/itex] can be characterised as follows:

The preimage of any open (resp. closed) set in [itex](Y,d_Y)[/itex] is open (resp. closed) in [itex](X,d_X)[/itex].

Hence, for example if we define [itex]f_1 (x,y) = x-y[/itex] then [itex]f_1[/itex] is continuous and [itex]A_1 = f_1^*\left( (-\infty ,1] \right)[/itex]. Since [itex](-\infty , 1][/itex] is closed, [itex]A_1[/itex] is closed.

Similarly for [itex]A_2[/itex] and [itex]A_3[/itex], but not sure about [itex]A_4[/itex]. Can I write it in a way that makes it more obvious/easier to work with the preimage?

If [itex]X,Y[/itex] are sets and [itex]f:X\to Y[/itex] is a function with [itex]B\subset Y[/itex], the

*preimage*is defined [itex]f^*(B) = \{x\in X : f(x) \in B\}[/itex].If [itex]d_X, d_Y[/itex] are metrics on [itex]X,Y[/itex], continuity of [itex]f[/itex] can be characterised as follows:

The preimage of any open (resp. closed) set in [itex](Y,d_Y)[/itex] is open (resp. closed) in [itex](X,d_X)[/itex].

Hence, for example if we define [itex]f_1 (x,y) = x-y[/itex] then [itex]f_1[/itex] is continuous and [itex]A_1 = f_1^*\left( (-\infty ,1] \right)[/itex]. Since [itex](-\infty , 1][/itex] is closed, [itex]A_1[/itex] is closed.

Similarly for [itex]A_2[/itex] and [itex]A_3[/itex], but not sure about [itex]A_4[/itex]. Can I write it in a way that makes it more obvious/easier to work with the preimage?

Last edited by a moderator: