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I have to show two metric functions $d_1$ and $d_2$, defined over the same set $X$, such as the identity function defined over $(X,d_1)$ in $(X,d_2)$ is not continuous. This has to be done with the negation of limit definition, i.e.

$$\exists\epsilon>0,\forall \delta>0(d_1(x,x_0)<\delta \land d_2(x,x_0)\geq\epsilon)$$

I guess that this can't be done because for all $\delta>0$ implies that $x$ can be equal to $x_0$, so $0=d_2(x,x_0)<\epsilon$ for all $\epsilon$. Then, it does not exist any $\epsilon$.

Julian
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    You have not negated the definition of continuity correctly. You forgot about the quantifier "$\forall x\in X$" in the definition which comes after $\exists \delta>0$. So the $x$ in your statement can depend on $\delta$, so you get no contradiction. – Eric Wofsey Dec 12 '16 at 02:30
  • http://math.stackexchange.com/questions/435104/is-the-identiy-function-continous-on-equivalent-metric-spaces –  Dec 12 '16 at 02:30
  • @EricWofsey You mean $\forall x \in X$ because the function is continuous in all points? Is in this way, why it does not goes at the begining? So the negation is

    $$\exists x\in X,\exists\epsilon>0,\forall \delta>0(d_1(x,x_0)<\delta \land d_2(x,x_0)\geq\epsilon)$$

    And this gets the same, we can select $x=x_0$ and its done

     – Julian Dec 12 '16 at 02:43
  • @user3161790: You still have it wrong. The negation of continuity at $x_0$ is $$\exists\epsilon>0,\forall\delta>0,\exists x\in X,\big(d_1(x,x_0)<\delta\land d_2(x,x_0)\ge\epsilon\big);.$$ And no, you clearly cannot take $x=x_0$, since $d_2(x_0,x_0)=0\not\ge\epsilon$. – Brian M. Scott Dec 12 '16 at 02:55
  • @BrianM.Scott: But, if $\forall \delta>0 (d_1(x,x_0)<\delta)$, I guess that $d_1(x,x_0)=0$, since it has to be lesser than all positive numbers (represented as $\delta$). Since $d_1$ is a metric space, $x=x_0$. – Julian Dec 12 '16 at 03:02
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    @user3161790: No, no, no. You’re reading the expression incorrectly. It absolutely does not say that $d_1(x,x_0)<\delta$ for each $\delta>0$. It says that there is some fixed positive $\epsilon$ such that no matter what positive $\delta$ you choose, there is some $x\in X$ (whose identity may depend on $\delta$) such that on the one hand $d_1(x,x_0)<\delta$, but on the other hand $d_2(x,x_0)\ge\epsilon$. It can be a different $x$ for each $\delta$. – Brian M. Scott Dec 12 '16 at 03:08
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    @BrianM.Scott I think I get it. Let $X=\mathbb{R}$, $d_1(x,x_0)=|x-x_0|$ and $d_2$ the discrete metric. Let us suppose that $|x-x_0|=\delta /2$ with $x>x_0$, so $x=x_0+\delta /2$.

    Select $\epsilon = 1/2$, then, $d_2(x,x_0)=d_2(x_0+\delta /2, x_0)=1$ since $x\neq x_0$. We have shown that exists some $x\in X$ such as $d_2(x,x_0)=1\geq \epsilon=1/2$

    Is this well done? Is there a way to generalize this to any set $X$?

     – Julian Dec 12 '16 at 03:37
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    @user3161790: Yes, that example will work just fine. In fact you can use any positive $\epsilon\le 1$: no matter what $\delta>0$ you pick, you can let $x=x_0+\frac{\delta}2$ and have $d_1(x,x_0)<\delta$ and $d_2(x,x_0)=1\ge\epsilon$. – Brian M. Scott Dec 12 '16 at 03:41

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