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Let $(X,d)$ and $(Y,e)$ be metric spaces , and let $f: X \to Y$ be a function.

True or false ? Give a proof or a counterexample as appropriate.

$(a)$ If $d$ is the discrete metric on $X$,then $f$ is continuous.

$(b)$ If $U$ open in $(X,d) \Rightarrow f(U)$ open in $(Y,e)$, then $f$ is continuous.

$(c)$ If there exists $r\gt 0$ so that $\,\,\,e(f(x_1),f(x_2)) \le r\,d(x_1,x_2)\,\,$for every $x_1,x_2 \in X$ , then $f$ is continuous. (Dose the converse hold?)

Definition:

  • Let $(X, d)$ and $(Y, e)$ be metric spaces, and let$ x \in X$. A function $f : X \to Y $is continuous at $x$ if: $\forall B \in \mathcal B(f(x)) \exists A \in \mathcal B(x) : f(A) \subseteq B$
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  • $\begingroup$ What are your thoughts on the given problem ? $\endgroup$ – Belgi Apr 13 '13 at 18:07
  • $\begingroup$ @Belgi: for (a) I think it is true because every open ball in Y there is an open ball in X with the discrete metric. since all the sets in the discrete metric is open and closed in the same time. for (b) I think is false but I couldn't think of a counterexample. for (c) I don't have an answer. $\endgroup$ – Jhwana Apr 13 '13 at 18:26
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HINTS:

(a) If $d$ is the discrete metric on $X$, what are the open sets in $X$?

(b) Try thinking of a metric on $Y$ that gives a lot of open sets (so that the condition will hold). Working out (a) will tell you the metric. Then think about the function (so that it is not continuous).

(c) Think of the "$\epsilon -\delta$" definition of continuity.

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  • $\begingroup$ for (a) the open sets are $X$ itself and $\{x\}$ for all $x \in X$ $\endgroup$ – Jhwana Apr 13 '13 at 18:25
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    $\begingroup$ Yes. So what $A$ do you choose that fulfills the definition? $\endgroup$ – Weltschmerz Apr 13 '13 at 18:28
  • $\begingroup$ $A=\{x\}$...Thank you $\endgroup$ – Jhwana Apr 13 '13 at 18:31
  • $\begingroup$ for(b) If $(X,d)$ is the discrete metric on $X$ and $(Y,e)=(\Bbb R,d_{\Bbb R})$ and $f(x)= \begin{cases} 1, & \text{if $x \ge 0$} \\ 0, & \text{if $x \lt 0$} \\ \end{cases}$ ,Does this work? $\endgroup$ – Jhwana Apr 13 '13 at 18:41
  • $\begingroup$ Well, you want "a lot" of open sets in $Y$, not in $X$... $\endgroup$ – Weltschmerz Apr 13 '13 at 18:45

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