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Oxymoron

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There are several definitions of a continuous function between metric spaces. Let [itex](X,d_X)[/itex] and [itex](Y,d_Y)[/itex] be metric spaces and let [itex]f:X\rightarrow Y[/itex] be a function. Then we have the following as definitions for continuity of [itex]f[/itex]:

[tex]\square \quad \forall\, x \in X \mbox{ and } \forall \, \epsilon >0 \, \exists \, \delta > 0 \mbox{ such that } d_X(y,x) < \delta \, \Rightarrow \, d_Y(f(y),f(x)) < \epsilon [/tex]

[tex]\square \quad \forall\, x \in X \mbox{ and } \forall \, \epsilon >0 \, \exists \, \delta > 0 \mbox{ such that } f(B(x,\delta)) \subseteq B(f(x),\epsilon)[/tex]

[tex]\square \quad \mbox{For every open subset } U \mbox{ of } Y \mbox{ the inverse image } f^{-1}(U) \mbox{ is an open subset of } X[/tex]

I want to prove that each of these statements are equivalent to each other. When I do this, is it okay to prove (1) <=> (2), (2) <=> (3), and (3) <=> (1), that is, I dont have to prove (2) => (1) once I have proven (1) => (2).

Secondly, is it okay if I just prove that each statement proves that [itex]f[/itex] is continuous and hence they are equivalent (ie. the statements are equivalent because they each prove continuity of [itex]f[/itex]?). If this is not the right way to prove that each statement is equivalent, could someone tell me how I would do that.

[tex]\square \quad \forall\, x \in X \mbox{ and } \forall \, \epsilon >0 \, \exists \, \delta > 0 \mbox{ such that } d_X(y,x) < \delta \, \Rightarrow \, d_Y(f(y),f(x)) < \epsilon [/tex]

[tex]\square \quad \forall\, x \in X \mbox{ and } \forall \, \epsilon >0 \, \exists \, \delta > 0 \mbox{ such that } f(B(x,\delta)) \subseteq B(f(x),\epsilon)[/tex]

[tex]\square \quad \mbox{For every open subset } U \mbox{ of } Y \mbox{ the inverse image } f^{-1}(U) \mbox{ is an open subset of } X[/tex]

I want to prove that each of these statements are equivalent to each other. When I do this, is it okay to prove (1) <=> (2), (2) <=> (3), and (3) <=> (1), that is, I dont have to prove (2) => (1) once I have proven (1) => (2).

Secondly, is it okay if I just prove that each statement proves that [itex]f[/itex] is continuous and hence they are equivalent (ie. the statements are equivalent because they each prove continuity of [itex]f[/itex]?). If this is not the right way to prove that each statement is equivalent, could someone tell me how I would do that.

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