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Let $X := (0,\infty)$ and $\left(X, \rho: X\times X\rightarrow (0,\infty), (x,y)\mapsto\rho(x, y):=\left|\frac{1}{x}-\frac{1}{y}\right|\right)$ a metric space.

Determine whether it is complete or not.

I am aware of the following possibilities:

  1. Show that any cauchy sequence converges
  2. Find a cauchy sequence that does not converge
  3. Find a space which $(X,\rho)$ is a closed subspace of and look if cauchy sequences in $X$ converge in this space

For option 3, I can't find a proper space.

For options 1 and 2, I do not know how to determine how the limit should look like to test if a cauchy sequence converges to it, or I can't think of any “special” function that would be cauchy, but be divergent.

I tried to take a look at functions like $f(x):=\frac{1}{x}$, because they tend to go to zero (at least with the standard metric), and $0\not\in X$, but they weren't even cauchy.

Which approach should I take? Is there something general or something like a “special trick” in this case?

EDIT: Here my solution

The concept was that $(a_n)_{n\in\mathbb{N}}:=n$ is cauchy, but not convergent in X — rudimentary said, because $\infty\not\in X=(0,\infty)$.

Lemma: $a_n$ is cauchy in $X$.

Proof: Let $\varepsilon>0, N:=\lceil\frac{1}{\varepsilon}\rceil, n,m>N$, without loss of generality $n≥m>0$. Then $$ \rho(a_n,a_m)=\left|\frac{1}{n}-\frac{1}{m}\right|=\frac{1}{m}-\frac{1}{n}<\frac{1}{m}<\frac{1}{N}≤\varepsilon $$ Thus, $\forall \varepsilon>0\quad\exists N\in\mathbb{N}\quad\forall\ m,n>N:\rho(a_n,a_m)<\varepsilon$ q.e.d.

Similiar, we can construct another lemma, which can informally be interpreted as “$a_n\rightarrow\frac{1}{0}$”:

Lemma 2: $\forall\varepsilon>0\exists N\in\mathbb{N}\forall n>N:\left|\frac{1}{n}-0\right|<\varepsilon$

Proof: Let $\varepsilon>0, N:=\lceil \frac{1}{\varepsilon}\rceil,n>N$ Then $$\left|\frac{1}{n}-0\right|=\frac{1}{n}<\frac{1}{N}<\varepsilon$$q.e.d.

Theorem: $(X,ρ)$ is not complete.

Proof:

assume $\exists x\in X: a_n\rightarrow x$. This is equivalent to $\forall\varepsilon>0\exists N_1\in\mathbb{N}\forall n>N_1:\left|\frac{1}{n}-\frac{1}{x}\right|<\varepsilon$.

Using lemma 2, we get: $$ ⇔ \forall \varepsilon>0\exists N_2\in\mathbb{N}\forall\ n>N_2:\left|\frac{1}{n}-0\right|<\varepsilon\\ \wedge \forall\varepsilon>0\exists N_1\in\mathbb{N}\forall n>N_1:\left|\frac{1}{n}-\frac{1}{x}\right|<\varepsilon\\ ⇔ \forall \varepsilon>0\exists N:=\max\{N_1,N_2\}\in\mathbb{N}\forall n>N: \left( \left|\frac{1}{n}-0\right|<\varepsilon \wedge \left|\frac{1}{n}-\frac{1}{x}\right|<\varepsilon \right)\\ ⇒ \forall \varepsilon>0\exists N\in\mathbb{N}\forall n>N: \left|\frac{1}{x}-0\right|=\left|\frac{1}{x}-\frac{1}{n}+\frac{1}{n}-0\right|≤\left|\frac{1}{n}-\frac{1}{x}\right| + \left|\frac{1}{n}-0\right| <\varepsilon+\varepsilon=2\varepsilon $$

Because $\not\exists x\in X=(0,\infty):0=\frac{1}{x}=\left|\frac{1}{x}-0\right|⇒0<\frac{\left|\frac{1}{x}-0\right|}{2}$, we can choose this for $ε$: $$⇒\exists N\in\mathbb{N}\forall n>N:\left|\frac{1}{x}-0\right|<2ε=\left|\frac{1}{x}-0\right|\\ ⇔\bot\quad\text{(contradiction)}$$ Thus, the assumption is wrong and $a_n$ is divergent and cauchy in $X$ which makes $(X,ρ)$ not a complete metric space. q.e.d.

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    $\begingroup$ If approaching $0$ doesn't work, maybe you should look at the other end. $\endgroup$ – Daniel Fischer Jan 11 '15 at 17:10
  • $\begingroup$ Ah, I see the concept. So I take a look at the cauchy seqience $a_n=n$: $\forall \epsilon = 0\exists N\in\mathbb{N}\forall n,m>N: \left|\frac{1}{n}-\frac{1}{m}\right|<\epsilon$ (which I have to prove). Because I can choose $m$ to be arbitrarily high, I find $\forall \epsilon = 0\exists N\in\mathbb{N}\forall n>N: \left|\frac{1}{n}-0\right|<\epsilon$ (is that true?), which means there has to be a limit $l: \frac{1}{l}=0\Rightarrow l\not\in(0,\infty)$. Is this Idea correct? $\endgroup$ – Lukas Juhrich Jan 11 '15 at 17:26
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Try with the series $x_n=n$, which is Cauchy and converges to $ \infty $. This means that X is not complete.

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One approach: theorem: let $f: (X,d) \rightarrow (Y,\rho)$ be a bijective isometry between metric spaces. Then one is complete iff the other is.

The proof is straightforward: suppose $(X,d)$ is complete. Let $(y_n)$ be a Cauchy sequence in $(Y, \rho)$, so $\forall \epsilon > 0 \exists N \forall n,m \ge N : \rho(y_n, y_m) < \epsilon$. As $f$ is a bijection, we have (unique) $x_n \in X$ with $f(x_n) = y_n$. Then as $f$ is an isometry, $d(x_n, y_n) = \rho(f(x_n), f(x_m) = \rho(y_n, y_m)$ for all $n,m$, and so it follows that $(x_n)$ is a Cauchy sequence in $(X,d)$, and so it has a limit $x \in X$, so $\forall \epsilon > 0 \exists N \forall n \ge N: d(x_n, x) < \epsilon$. But taking $y =f(x)$ and using the isometry property (or even continuity of $f$ would suffice) we see that $(y_n) = (f(x_n))$ converges to $f(x) = y$. So $(Y, \rho)$ is complete. And if $(Y, \rho)$ is complete, take any Cauchy sequence $(x_n)$ in $(X,d)$. And then similarly $(y_n) = (f(x_n))$ is Cauchy in $(Y, \rho)$ by the isometry property, and so converges to some $y \in Y$. And taking the $x$ such that $f(x) = y$ we again see that $(x_n)$ converges to $x$.

Now, note that $f(x) = \frac{1}{x}$ is an isometry between $((0, \infty), |\cdot |)$ and $((0,\infty), \rho)$ almost by definition of $\rho$. And the first space is not complete (as $\frac{1}{n}$ is Cauchy (it's convergent in the larger $([0,\infty), |\cdot |)$ to $0$, and convergent sequences are Cauchy, and so Cauchy in any set that contains all sequence elements in the same metric), but not convergent (or it would have a limit in $(0,\infty)$ (!), but then this limit would also be a limit in the larger set again, which cannot be: it is there convergent to $0$ and limits of sequences are unique!).

So your space of interest is also not complete.

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