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Let $(X,d_{disc})$ be a discrete metric space, where $d_{disc}(x,x')=\begin{cases} 1 & \text{if } x\neq x', \\ 0 & \text{if }x=x'.\end{cases}$

Let $(Y,d_Y)$ be an arbitrary metric space, and $\mathbb{R}$ be equipped with $d_{\mathbb{R}}(r,r')=|r-r'|.$

  1. Prove that any function $f:X\to \mathbb{R}$ is continuous.
  2. Prove that $d_Y:Y\times Y\to \mathbb{R}$ is continuous. Here articulate which metric you use for $Y\times Y$.

My Idea:

  1. Observe that $f$ is continuous iff for all $\epsilon>0$, $\exists \delta>0$ such that $d_{\mathbb{R}}(r,r')<\epsilon$ if $d_{disc}(x,x')<\delta.$ Pick $\delta=1/2$ then if $x=x'$ trivially $d_{\mathbb{R}}(r,r')<\epsilon$ or else if $x\neq x'$ then $d_{\mathbb{R}}(r,r')=|r-r'|<\epsilon$? I don't understand how to finish it. Alternatively, as Nate said I could look at the other definition, $f$ is continuous iff forall open sets $U\subset \mathbb{R}$ , $f^{-1}(U)\underset{open}{\subset} X.$ We are concerned with all the open balls with radius less than $1$.
  2. I am not sure how to proceed on this one. Any comments and hints will be appreciated.Thanks.
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Your proof for (1) seems a bit too complicated. You have to show that for any $\epsilon > 0$ you can find a $\delta > 0$ such that if $d_{\text{disc}}(x,x') < \delta$ then $d_Y(f(x), f(x')) < \epsilon$. Given any $\epsilon$, you don't have to think very hard to find a $\delta$ which will work. (In fact, it won't even have to depend on $\epsilon$.) Alternatively, use the fact that $f$ is continuous if $f^{-1}(U)$ is open for every open $U$; what are the open sets in $X$? – Nate Eldredge Apr 1 '12 at 22:44
Thanks, I forgot this definition. I will post my revised solution soon. – Lyapunov Apr 1 '12 at 22:49
For (2), see this answer of Srivatsan's. – Dylan Moreland Apr 2 '12 at 2:08
your last comment on (1) is incorrect. Any union of open balls are open and around every point an open ball of radius one-half contains a single point. So any set is an open set. A ball of radius 2 is still open-it just happens to be the entire space(which better be open if we want to talk about the space's topology) – Jeremy Mann Apr 2 '12 at 2:09
For your revised 1: if $x \ne x'$ then $d(x,x') > \delta$... – Nate Eldredge Apr 2 '12 at 14:15

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