2
$\begingroup$

Give an example of a function which is continuous with respect to some metric but not continuous with respect to some other metric.

(I take the identity function from (R,d_1) to (R,d_2) where d_1 is discrete metric and d_2 is usual metric. It is continuous on R. I am trying to replace the metric d_1 only (not d_2) by some other metric, say d_3, so that the identity function from (R,d_3) to (R,d_2) becomes discontinuous.)

$\endgroup$

closed as off-topic by Crostul, martini, Claude Leibovici, Behrouz Maleki, Henrik Jan 7 '17 at 15:43

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Crostul, martini, Claude Leibovici, Behrouz Maleki, Henrik
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ Consider the discrete metric: in such a metric all functions are continuous! $\endgroup$ – Crostul Jan 7 '17 at 11:16
1
$\begingroup$

Let $X=\mathbb R$; consider metrics $d_1,d_2$ with $$d_1(x,y)=|x-y|$$ and $$d_2(x,y)=\begin{cases}1 &\text{if}\ x\ne y \\ 0 &\text{if}\ x=y\end{cases}$$ If $\iota:X\to X$ is the identity function, then $$\iota:(X,d_2)\to(X,d_1)$$ is continuous at all points $y\in X$. Given any $\varepsilon>0$, take $\delta=1$; then for any $x\in X$, $d_2(x,y)<\delta$ $\implies$ $x=y$ $\implies$ $d_1(x,y)=0<\varepsilon$.

On the other hand $$\iota:(X,d_1)\to(X,d_2)$$ is not continuous at any point $y\in X$. Given $\varepsilon=1$ and any $\delta>0$, we can always find $x$ such that $0<|x-y|=d_1(x,y)<\delta$; then $d_2(x,y)=1\ge\varepsilon$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.