Metric Spaces Homework: Showing Cauchy Sequences

[Ted123]

Homework Statement

Homework Equations

The Attempt at a Solution

I've done the first 3 parts. I've come to the bit on Cauchy sequences at the end. How do I show $x_n = n$ is/isn't a Cauchy sequence in the 2 metrics?

$(x_n)$ is a Cauchy sequence in a metric space $(X,d)$ if for any $\varepsilon >0$ there exists $N\in \mathbb{N}$ such that if $m,n > N$ then $d(x_m , x_n ) < \varepsilon$.

The metric space $(X,d)$ is complete if every Cauchy sequence in $(X,d)$ converges to a limit in $X$.

 

[micromass]

So let's prove that it isn't a Cauchy sequence in the standard metric.

So, we need to find an epsilon (let's take $\epsilon=1$ for a moment), such that for all N, there exists n,m>N such that

$$d(x_n,x_m)\geq 1$$

So we must find n,m>N such that

$$|n-m|\geq 1$$

Can you find such a n and m??

 

[radou]

Well, let ε be some number in <0, 1>. Does there exist some positive integer N such that for all m, n >= N you have |xm - xn| < ε?

Edit: sorry, micromass seems to have answered first.

 

[Ted123]

So let's prove that it isn't a Cauchy sequence in the standard metric.

So, we need to find an epsilon (let's take $\epsilon=1$ for a moment), such that for all N, there exists n,m>N such that

$$d(x_n,x_m)\geq 1$$

So we must find n,m>N such that

$$|n-m|\geq 1$$

Can you find such a n and m??

Setting $m=N+1$ and $n=N+2$ we have that $m,n > N$ and $$|m-n|=|N+1-(N+2)|=|1-2|=1 \geqslant 1\;,\;\text{for all}\;N\in \mathbb{N}$$

 

[micromass]

Good! So that proves that it isn't a Cauchy sequence.

Now, to prove that it is a Cauchy sequence in the other metric, you must make

$$|\tan^{-1}(n)-\tan^{-1}(m)|$$

smaller than $\varepsilon$

 

[Ted123]

Good! So that proves that it isn't a Cauchy sequence.

Now, to prove that it is a Cauchy sequence in the other metric, you must make

$$|\tan^{-1}(n)-\tan^{-1}(m)|$$

smaller than $\varepsilon$

How do I make $|\tan^{-1}(m) - \tan^{-1}(n)|<\varepsilon$ ? I'm guessing that the fact given in the question that $d(x,y)< \pi$ might help?

However, for the last part on whether $(\mathbb{R} ,d)$ is complete, can I say: if it is complete then every Cauchy sequence in $(\mathbb{R} ,d)$ must converge to a limit in $\mathbb{R}$.

Suppose the Cauchy sequence $x_n=n$ converges in $(\mathbb{R} ,d)$. Then by part (c), $x_n$ must converge in the standard metric. Every convergent sequence is Cauchy, but $x_n$ is not Cauchy in the standard metric - contradiction. So $x_n$ does not converge in $(\mathbb{R} , d)$; therefore not every Cacuhy sequence in $(\mathbb{R} , d)$ converges so it is not complete.

 

[micromass]

How do I make $|\tan^{-1}(m) - \tan^{-1}(n)|<\varepsilon$ ? I'm guessing that the fact given in the question that $d(x,y)< \pi$ might help?

Well, this showing that $x_n=n$ is Cauchy in $(\mathbb{R},d)$ is equivalent to showing that $y_n=\tan^{-1}(n)$ is Cauchy in the standard metric.

However, for the last part on whether $(\mathbb{R} ,d)$ is complete, can I say: if it is complete then every Cauchy sequence in $(\mathbb{R} ,d)$ must converge to a limit in $\mathbb{R}$.

Suppose the Cauchy sequence $x_n=n$ converges in $(\mathbb{R} ,d)$. Then by part (c), $x_n$ must converge in the standard metric. Every convergent sequence is Cauchy, but $x_n$ is not Cauchy in the standard metric - contradiction. So $x_n$ does not converge in $(\mathbb{R} , d)$; therefore not every Cacuhy sequence in $(\mathbb{R} , d)$ converges so it is not complete.

That is good.

 

[Ted123]

Well, this showing that $x_n=n$ is Cauchy in $(\mathbb{R},d)$ is equivalent to showing that $y_n=\tan^{-1}(n)$ is Cauchy in the standard metric.

In examples I've seen on showing a sequence is Cauchy they often involve getting $d(x_m , x_n) \leqslant X \to 0$ where $X$ is some upper bound. Why does this show the sequence is Cauchy?

For my sequence $x_n = n$ I could use the triangle inequality $d(x_m , x_n) \leqslant d(x_m , \frac{\pi}{2} ) + d(\frac{\pi}{2} , x_n)$: $$d(x_m ,x_n) = | \tan^{-1} (x_m) - \tan^{-1}(x_n) |$$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \leqslant | \tan^{-1}(m) - \frac{\pi}{2} | + | \tan^{-1}(n) - \frac{\pi}{2} | \to 0 + 0=0$ as $m,n \to \infty$

since $\displaystyle \lim_{r\to\infty} \tan^{-1}(r) = \frac{\pi}{2}$

 

[micromass]

In examples I've seen on showing a sequence is Cauchy they often involve getting $d(x_m , x_n) \leqslant X \to 0$ where $X$ is some upper bound. Why does this show the sequence is Cauchy?

Well, write out what it means that $d(x_m,x_n)\rightarrow 0$...

For my sequence $x_n = n$ I could use the triangle inequality $d(x_m , x_n) \leqslant d(x_m , \frac{\pi}{2} ) + d(\frac{\pi}{2} , x_n)$: $$d(x_m ,x_n) = | \tan^{-1} (x_m) - \tan^{-1}(x_n) |$$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \leqslant | \tan^{-1}(m) - \frac{\pi}{2} | + | \tan^{-1}(n) - \frac{\pi}{2} | \to 0 + 0=0$ as $m,n \to \infty$

since $\displaystyle \lim_{r\to\infty} \tan^{-1}(r) = \frac{\pi}{2}$

That's good.

 

[Office_Shredder]

For my sequence $x_n = n$ I could use the triangle inequality $d(x_m , x_n) \leqslant d(x_m , \frac{\pi}{2} ) + d(\frac{\pi}{2} , x_n)$: $$d(x_m ,x_n) = | \tan^{-1} (x_m) - \tan^{-1}(x_n) |$$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \leqslant | \tan^{-1}(m) - \frac{\pi}{2} | + | \tan^{-1}(n) - \frac{\pi}{2} | \to 0 + 0=0$ as $m,n \to \infty$

Be careful when you're writing this up! You're using the triangle inequality for absolute value here, but you're not using the triangle inequality for the metric d here - if you were actually using d(x_n,pi/2) you would end up with |arctan(n)-arctan(pi/2)|+|arctan(m)-arctan(pi/2)| which doesn't go to zero as n and m go to infinity

 

[Ted123]

Be careful when you're writing this up! You're using the triangle inequality for absolute value here, but you're not using the triangle inequality for the metric d here - if you were actually using d(x_n,pi/2) you would end up with |arctan(n)-arctan(pi/2)|+|arctan(m)-arctan(pi/2)| which doesn't go to zero as n and m go to infinity

I'm not getting any credit for this; it's all past exam paper stuff.

But, yes - it's the scalar triangle inequality I should be using!