0

Let $X=\{0,1\}^{N}$ and define for $x = (x_1, \dots x_N), \ y = (y_1, \dots, y_N)$

$$ d(x,y) = \sum_{n=1}^{N} [(x_n + y_n) \text{mod} 2] $$

Is this space complete?

My intuition says no. So I tried to come up with a counter example. If we have $N=1$ then the possibilities we have is $x=(1), \ x = (0)$ then the only possibilities we have for a Cachy sequence $(x_n)_{n \in \mathbb{N}}$ is when $x_n = x_m$, $m \not = n$

Olba12
  • 2,579

1 Answers1

1

Note that for any two elements $x,y \in X$, we either have $d(x,y) = 0$ (and hence $x=y$) or $d(x,y) \ge 1$. Hence a Cauchy sequence $(x_n)$ must have $$\forall\varepsilon > 0\; \exists n_0 \in \mathbf N \; \forall n,m \ge n_0 : d(x_n, x_m) < \varepsilon $$ there is some $n_0$ such that $x_n = x_m$ for $n,m \ge n_0$ (choose $\varepsilon = \frac 12$). As such eventually constant sequences are convergent, $X$ is complete.

martini
  • 84,101
  • ...for $n,m \geq N$.. Do you mean the $N$ specified in the problem or some natural number $N$? – Olba12 Nov 15 '16 at 11:14
  • 1
    I just forgot that there is an $N$ is the question, will change my answer accordingly. – martini Nov 15 '16 at 11:15
  • By $\epsilon = \frac{1}{2}$ do you mean, that if we choose such an epsilon and we find a sequence such that $d(x_n,x_m) < \epsilon$ than we know for sure it is Cauchy? – Olba12 Nov 15 '16 at 11:22
  • Now, given any Cauchy-sequence, we use the definition with $\varepsilon = \frac 12$. – martini Nov 15 '16 at 11:23
  • the definition for convergence? I'm sorry, but I do not understand what you want to say with that. – Olba12 Nov 15 '16 at 11:25
  • The definition for being Cauchy, namely $$ \forall \epsilon > 0; \exists n_0 \in \mathbf N; \forall n,m \ge n_0: d(x_n,x_m) < \epsilon $$ – martini Nov 15 '16 at 11:27
  • I understand that my question was unclear. Given that $(x_n)$ is Cauchy, then for all $\epsilon > 0$ specially $\epsilon = 1/2$ we have that $\exists n_0 \in \mathbb{N}$ such that $\forall n,m > n_0$ $d(x_n,x_m) < 1/2$ and thus convergence? Is that what you are saying? – Olba12 Nov 15 '16 at 11:37