A not complete metric space?

Question

Please ,how to prove that the space $\mathbb{R}$ endowed with the metric $d(x,y)=|e^x-e^y|$ is not a complete space?

I don't find a Cauchy sequence but not convergent

Please

Thank you.

Can you think of a sequence $(x_n)$ where the Cauchy difference $d(x_n, x_m)$ gets small even though the $x_n$ themselves do not? — Simon S, Dec 24 '14 at 16:32
We know that $\mathbb{R}$ with the usual metric $d(x,y) =|x-y|$ is a complete metric space. Since the exponential function is continuous, then we usually would expect the $|e^{x}-e^{y}|$ should be small aswell... But maybe it has something to so with uniformness.. — TheOscillator, Dec 24 '14 at 16:46
@TheOscillator: No, continuity of the exponential doesn't suggest that the new metric is complete, and in fact it isn't. — Brian M. Scott, Dec 24 '14 at 16:51
Related (duplicate?): http://math.stackexchange.com/questions/153707/checking-for-completeness-of-mathbbr-with-metric-defined-by-d-1x-y-mi — Martin Sleziak, Dec 24 '14 at 20:38
@TheOscillator: If $e^x$ were a slowly-growing function (e.g. with $e^x<x$) we might expect that $|x-y|$ small implies $|e^x-e^y|$ small. But we know that actually it is a very quickly-growing function; it's not obvious that this means $|e^x-e^y|$ is not small, but it lends some plausibility to the claim. (I don't know if you were intending your comments as a pedagogical device, but either way this may be useful to the OP as well.) — Eric Stucky, Jan 03 '15 at 03:17
As Martin suggests, this is a duplicate of that question which is answered in the comments. — Eric Stucky, Jan 03 '15 at 03:21

TorsionSquid · Accepted Answer · 2014-12-24T16:54:45.743

20

Consider the sequence $x_n=-n$. Clearly it doesn't converge. For, suppose $x_n \to x$ for some real number $x$. Then $|e^{-n}−e^x|→0$. Hence, $e^x=0$, which cannot be.

However, $N<n<m \implies |e^{x_n}-e^{x_m}|=e^{-n}-e^{-m}<e^{-N} \to 0$ as $N \to \infty$. So the sequence is Cauchy.

edited Dec 24 '14 at 16:54

answered Dec 24 '14 at 16:38

TorsionSquid

3,530

how you see that it soesn't converge with this metric ? – Vrouvrou Dec 24 '14 at 16:42
@Vrouvrou whether or not a sequence converges doesn't depend on the metric. Whether or not it is a Cauchy sequence does. – Matt Samuel Dec 24 '14 at 16:43
you just do $\lim_{n\rightarrow +\infty} -n=-\infty $ that's all ? – Vrouvrou Dec 24 '14 at 16:44
1

Suppose $x_n \to x$ for some real number $x$. Then $|e^{-n}-e^x| \to 0$. Hence, $e^x=0$, which cannot be. So the sequence doesn't converge with the given metric. – TorsionSquid Dec 24 '14 at 16:45
7

@MattSamuel Uh? Whether a sequence converges or not is highly dependent on the metric. See this for example. – Najib Idrissi Dec 24 '14 at 16:45
@NajibIdrissi if two metrics induce the same topology, then they have the same convergent sequences. Convergence depends only on the topology. – Matt Samuel Dec 24 '14 at 16:46
i have a defiition we say that sequence is convergent if : $\forall \varepsilon>0, \exists n_0\in \mathbb{N}, \forall n\geq n_0, d(x_n,x)<\varepsilon$ – Vrouvrou Dec 24 '14 at 16:47
3

@MattSamuel Yes, but the topology depends on the metric... You would need to prove that this new metric induces the usual topology on $\mathbb{R}$. – Najib Idrissi Dec 24 '14 at 16:47
@NajibIdrissi looking at your example, notice that the accepted answer says that the metric given there does not induce the usual topology on the real numbers. – Matt Samuel Dec 24 '14 at 16:47
The sequence is Cauchy. It does not converge to any real number, because such a real number would need to satisfy $e^x=0$. – Ian Dec 24 '14 at 20:43
juste why $e^{-n}-e^{-m}<e^{-N}$ please – Vrouvrou Jan 12 '15 at 21:16
Note that $e^{-x}$ is a positive, decreasing function. – TorsionSquid Jan 13 '15 at 20:01

score 14 · Answer 2 · answered Dec 24 '14 at 19:05

Here is a less direct way of viewing the problem. The function $f(x) = e^x$ defines an isometry from your metric space to the set $\mathbb{R}^+$ of positive real numbers with the usual metric. This fact is straightforward: $f$ is a bijection (the inverse function is the natural logarithm), and if $d$ is your metric and $d_1$ is the usual metric, then for $x,y \in \mathbb{R}$, we have $$d(x,y) = |e^x - e^y| = |(f(x) - f(y)| = d_1(f(x),f(y)).$$

So, $(\mathbb{R}, d)$ and $(\mathbb{R}^+, d_1)$ are isomorphic as metric spaces. We can see that the latter space is not complete by considering any sequence in $\mathbb{R}^+$ converging to 0. Such a sequence is Cauchy, but does not converge to a point of $\mathbb{R}^+$.

Nice answer! I think it makes it even more clear why we should consider a Cauchy sequence tending to $-\infty$. — Viktor Vaughn, Dec 24 '14 at 19:20

score 7 · Answer 3 · answered Dec 24 '14 at 16:41

7

HINT: Is $d(-100,-101)$ a large, medium, or small number? Look at the graph of $y=e^x$.

answered Dec 24 '14 at 16:41

Brian M. Scott

616,228

@Vrouvrou: That's right. And what about $d(-100,n)$ for integers $n<-101$? – Brian M. Scott Dec 24 '14 at 16:47
@Vrouvrou: Now try to use that observation to find a really simple sequence of negative integers that diverges even though it's $d$-Cauchy. – Brian M. Scott Dec 24 '14 at 16:53

A not complete metric space?

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