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Is it possible to define a metric on $\mathbb R$ such that $(1,0,1,0,...)$ converges on $(\mathbb R, d)$?

I believe it is impossible. But how to show analytically? Any hint would be appreciated.

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Metric spaces are Hausdorff, since $0,1$ are distinct points, any sequence converging to $1$ eventually can no longer be near $0$. –  Adam Hughes Jul 23 at 2:10
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You would have (for some $x$) $d(1,x)=0$ and $d(0,x) = 0$ from which we would get $d(0,1) = 0$. –  copper.hat Jul 23 at 2:16

7 Answers 7

Assume that for a certain metric $d$ your sequence converges to $g\in\mathbb{R}$. Then its even subsequence also converges to $g$ but it converges to $0$ thus $g=0$ similarly considering odd subsequence one shows that $g=1$ which gives a contradiction.

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I think no.

Say distance is d, then forall $N$ you have d distance between members, so no N for $n>N$ shorter than thje distance $\epsilon$. So it not convergent.

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In greater detail than my comment. A metric space is Hausdorff, so given $x_n\to 1$ you know for $n>N(\epsilon)$ that $d(1,x_n)<\epsilon$, so choose

$$\epsilon = {d(0,1)\over 2}>0$$.

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Hint: Look at the odd/even subsequences. What do we know about limits of subsequences of a convergent sequence?

Alternate hint: Is it Cauchy?

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If the sequence converges, then any subsequence of it converges to the same limit. This can be proven easily for any metric space. Any constant sequence converges to its repeated element. This too holds in any metric space. It now follows that the sequence you suggest can never converge under any metric on $\mathbb R$. More is true. Given any alternating sequences (or more generally, any sequence that contains two distinct constant subsequences), then under no metric on the ambient set will the sequence converge.

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Let the $n$th term of our sequence be $a_n$. Suppose there is a metric $d$ such that $\{a_n \}$ converges on $(\mathbb{R}, d)$ to $p \in \mathbb{R}$. Since $0 \neq 1$, it follows from the definition of a metric that $d(1,0) > 0$. Set $\delta < d(1,0)/2$. Then there exists an $N$ such that for all $n \geq N$ we have $d(a_n,p) <\delta$. Assume WLOG that $a_N = 1$ (if not pick $a_{N+1}$). Then $a_{N+1} = 0$, and thus $$ d(1,0) < d(1,p) + d(p,0) = d(a_{N},p) + d(p, a_{N+1}) < \delta/2 + \delta/2 = \delta < d(1,0). $$ Hence we have arrived at a contradiction.

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You can also prove this is impossible working directly from definitions. Let $s_n$ be the sequence defined by $s_n=0$ if $n$ is even and $s_n=1$ if $n$ is odd. Suppose that this sequence converges to some $y$ with respect to some metric $d$. Then for all $\delta>0$, there exists $n$ so that $m>n$ implies $d(s_n,y)<\delta$.

Either $y \neq 0$ or $y \neq 1$. Since the argument is the same in either case, assume $y\neq 1$. Then $d(1,y)>0$, since one of the axioms for a metric is that $d(a,b)>0$ for $a\neq b$. Choosing $\delta<d(1,y)$ yields a contradiction since $d(s_m,y)>\delta$ for any odd $m\in \mathbb N$.

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If fact you would have $d(1,y) = d(0,y) = 0$. –  copper.hat Jul 23 at 2:17

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