Is $(X,d)$ is complete? Yes/no

Question

Given the metric space $(X,d)$ with $X := (0,\infty)$ and $d(x,y)=|\ln(x)-\ln(y)|$,

Is $(X,d)$ is complete?

My attempt : N0 . Let $(x_n) $ be a Cauchy sequence then $d(x_n, x_m)<\epsilon$ so $\mid \ln(x_n)-\ln(x_m)\mid \rightarrow 0.$ And $$d(x_n,x_m)=\mid \ln(x_n)-\ln(x_m)\mid=\mid \ln\frac{x_n}{x_m}\mid=0$$ So $\frac{x_n}{x_m} \rightarrow 1$ then subsequencec $(x_n)$ and $(x_m) $ have same limit and the sequence is convergent. But $d(x_n, x_m) = 0 \notin X$

So $(X, d)$ is not complete

Is its true ?

Answer 1

The metric space is complete. One of the major mistakes that caught my attention was when you wrote

$$d(x_n,x_m)=\mid \ln(x_n)-\ln(x_m)\mid=\mid \ln\frac{x_n}{x_m}\mid=0$$

instead of writing

$$d(x_n,x_m)=\mid \ln(x_n)-\ln(x_m)\mid=\mid \ln\frac{x_n}{x_m}\mid<\epsilon$$

as $=0$ does not mean $<\epsilon$.

Answer 2

There are many problems with your approach.

First, you do not spell out exactly what it means to be a Cauchy sequence.

Second, you write

$\mid \ln(x_n)-\ln(x_m)\mid \rightarrow 0$

but what is this supposed to mean exactly?

Third, you write

then subsequence $(x_n)$ and $(x_m) $ have

But what subsequences are you talking about?

Fourth, you write

But $d(x_n, x_m) = 0 \notin X$

Two problems with that. First, it is not clear what exactly you mean. Yet more crucially it is not really relevant where the distance which "is" $0$ is in the space or not. To appreciate this consider the space $[3,5]$ with the usual metric. I assume you know that it is complete, but your argument might show that it is not complete.

Finally, the space is actually complete.

Answer 3

Besides the logical errors in your proof attempt, as noted in the answers by "Axion004" and "quid♦", I'd like to point out that your approach was flawed from the outset.

It appears you were trying to show that every Cauchy sequence fails to converge, but that could never work, since, for example, in any metric space, a constant sequence is always a Cauchy sequence, and always converges.

For a given metric space, to prove it is not complete, it suffices to find one Cauchy sequence which doesn't converge. But that plan would fail for the metric space $(X,d)$ in question since it is, in fact, complete.

To show that $(X,d)$ is complete, we can argue as follows . . .

Suppose $(x_n)$ is a Cauchy sequence in $(X,d)$.

Our goal is to show that $(x_n)$ converges in $(X,d)$.

Let $(y_n)$ be the sequence in $\mathbb{R}$ defined by $y_n = \log(x_n)$.

Let $\epsilon > 0$.

Since $(x_n)$ is a Cauchy sequence in $(X,d)$, there exists a positive integer $N$ such that for all $m,n\ge N$, we have $|\log(x_m)-\log(x_n)|<\epsilon$.

Equivalently, for all $m,n\ge N$, we have $|y_m-y_n|<\epsilon$.

Hence $(y_n)$ is a Cauchy sequence in $\mathbb{R}$.

But then since $\mathbb{R}$ is complete, it follows that $(y_n)$ converges in $\mathbb{R}$ to $r$ say.

Using limits in $\mathbb{R}$, \begin{align*} &\lim_{n\to\infty}y_n=r\\[4pt] \implies\;&\lim_{n\to\infty}\log(x_n)=r\\[4pt] \implies\;&\lim_{n\to\infty}\log(x_n)=\log(e^r)\\[4pt] \implies\;&\lim_{n\to\infty}|\log(x_n)-\log(e^r)|=0\\[4pt] \implies\;&\lim_{n\to\infty}d(x_n,e^r)=0\\[4pt] \end{align*} Hence $(x_n)$ converges in $(X,d)$ to $e^r$.

It follows that $(X,d)$ is complete.