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 &gt; 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)|&lt;\varepsilon ? I'm guessing that the fact given in the question that d(x,y)&lt; \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)|&lt;\varepsilon ? I'm guessing that the fact given in the question that d(x,y)&lt; \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!