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Consider the space $(M,d)$ of sequences $x=(x_j :\ j\in\Bbb N),\ x_j\in A~\forall j$, where $A$ is a set, with $$d(x,y)=\begin{cases} \frac{1}{\min\{j\in\Bbb N^*\ :\ x_j\ne y_j\}} &\text{if}~~~ x\ne y \\0 &\text{if}~~~x=y \end{cases}$$

Question: prove that $(M,d)$ is a complete metric space.

My attempt:

Suppose a sequence $(x^{(n)})\subset M$ satisfies $\forall\epsilon>0~\exists N\in\Bbb N~\forall m,n\ge N~d(x^{(n)},x^{(m)})\le\epsilon$ i.e. $\frac{1}{\min\{j\in\Bbb N^*~:x^{(n)}_j\ne x^{(m)}_j\}}\le\epsilon$

I know there is a method where we do an analogy between our sequence and a sequence in $(\Bbb R,|\_|)$, which is complete, and the property of convergence can be passed on to $(x^{(n)})$ but I need a hint

John Cataldo
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2 Answers2

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The idea is to understand what $x^{(n)} \to x$ means in this space, where $x^{(n)} \in M$ is a sequence , and $x \in M$.

What it means is this : for all $\epsilon > 0$ there exists $N$ such that $n > N$ implies $d(x^{(n)},x) < \epsilon$. In other words, for all $n > N$ and $m < \frac{1}{\epsilon}$, we have $x_m^{(n)} = x_m$.

Now, I will leave you with points to fill.

  • Suppose $x^{(n)}$ is a Cauchy sequence.

  • Fix $k\in N$. Show that $x_k^{(n)}$ is an eventually constant sequence(in $n$) using the definition above. (A sequence is said to be eventually constant if there exists $l$ so that every term of the sequence is $l$ after some point)

  • Call the constant which $x_k^{(n)}$ converges to, as $x_k$. Now, using $x_k$ we define a sequence $x$.

  • Show that $x \in M$ and that $x^{(n)} \to x$.

Sarvesh Ravichandran Iyer
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  • I think it would be preferable if you would use the notation from John Cataldo for the double indices. – supinf Dec 10 '18 at 17:06
  • I will change the notation shortly. There, done. – Sarvesh Ravichandran Iyer Dec 10 '18 at 17:07
  • Why should the sequence become constant in $n$ after a while? – John Cataldo Dec 11 '18 at 13:40
  • Let $x^{(n)}$ be a Cauchy sequence. Fix $k \in \mathbb N$. Then, take $\epsilon = \frac 1k$. By definition of Cauchy, there exists $N$ such that $m,n \geq N$ implies $d(x^{(m)}, x^{(n)}) < \epsilon = \frac 1k$. Now, $d(x^{(m)},x^{(n)})$ is the reciprocal of the first position at which $x^{(m)}$ and $x^{(n)}$ differ. If this is smaller than $\frac 1k$, then that means that $x^{(m)}$ and $x^{(n)}$ don't differ at the $k$th position, but rather some position after it. Now, since this is true for $m,n \geq N$, we see that if $n >N$ then the $k$th position of $x^{(n)}$ remains constant. – Sarvesh Ravichandran Iyer Dec 11 '18 at 13:47
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hint:

If two elements $x,y$ are close (with respect to the metric $d$) this means that the first few coefficients $x_j,y_j$ are the same.

using the definition of a Cauchy sequence, it can be seen that for each $k$ there is an $N=N(k)$ such that for all $n\geq N(k)$ we have

$$ x_j^{(n)} = x_j^{(N(k))} \quad \text{for all}; j=1,\dots,k. $$

Note that $N(k)$ can be chosen such that $N:\mathbb N\to\mathbb N$ is an increasing function.

Then the limit point of $(x^{(n)})$ will be $$ y:= (y_k : k\in\mathbb N) := ( x_k^{N(k)} ). $$ It remains to show that $(x^{(n)})$ actually converges to $x$.

supinf
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