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Let $M$ denote the space of sequences $(x_n)$ where $x_n \in\{0,1\}$ for each $n$. Let $$d\colon M\times M\rightarrow\mathbb{R}\colon ((x_n),(y_n))\mapsto\sum_{i=1}^{\infty}|x_i-y_i|2^{-i}$$ be the usual sequence-space metric.

i) Let $U_0$ denote the set of sequences that begin with $0$. Show that $U_0$ is open.

ii) Show that $M$ is complete.

I think the first part requires something to do with uniform convergence under the metric, but I'm not sure and I can't get my head around an open set of sequences.

Dan Rust
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Tom
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2 Answers2

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Hint for the first part: show that if $(x_n)$ has $x_1=0$ and $(x_n')$ has $x_1'=1$, then $d((x_n),(x_n'))\geq\frac{1}{2}$. Given this, show that for every $s\in U_0$ we have $B_{\frac{1}{4}}(s)\subset U_0$.

Hint for the second part: If $(x_i)^{(j)}$ is a Cauchy sequence of elements in $M$, then show that for all $i$, there exists an $N\geq 1$ such that $x_i^{(j)}=x_i^{(N)}$ for all $j\geq N$.

Dan Rust
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  • could you add a little more detail - i'm still confused – Tom Oct 27 '13 at 23:57
  • What part are you confused about? Can you prove the parts in which I said 'show that'? If you can, they should give you a pretty big chunk of the proof for each part. – Dan Rust Oct 27 '13 at 23:58
  • i'm stuck on the second show that of the first part, and i don't see how it links together to show that U0 is open – Tom Oct 27 '13 at 23:59
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    Recall that a set in a metric space is open if for all elements in that set, there exists an open ball containing that element which is fully contained in the set. I'm suggesting that such an open ball can always be taken with any radius $<\frac{1}{2}$ (I chose $\frac{1}{4}$ just because it's a power of $2$.) because of what was shown in the first 'show that'. – Dan Rust Oct 28 '13 at 00:02
  • okay thanks I still can't work out how to show the second 'show that' of the first part though – Tom Oct 28 '13 at 00:19
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    Let $s'\in B_{\frac{1}{4}}(s)$. Suppose that $s'\notin U_0$ and so $s'{(1)}=1$. Then $d(s,s')\geq\frac{1}{2}$ by the first part, and so $s'\notin B{\frac{1}{4}}(s)$ which is a contradiction. It follows that $s'\in U_0$ and so $B_{\frac{1}{4}}(s)\subset U_0$ as required. – Dan Rust Oct 28 '13 at 00:23
  • Thank you so much - I understand so clearly now. Could you explain your notation in the second part please – Tom Oct 28 '13 at 00:36
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    $(x_i)^{(j)}$ is the $j$-th sequence in the Cauchy sequence of sequences indexed by $i$. $x_i^{(j)}$ is the $i$-th component of the $j$-th sequences of sequences. If you wrote the components of the sequences out in a $2$-d array, $i$ would be the horizontal component, $j$ would be the vertical. – Dan Rust Oct 28 '13 at 00:39
  • this bit is really confusing me - would you mind showing me a full proof that i can try to understand? – Tom Oct 28 '13 at 00:54
  • Given that I've laid out a full proof of the first part, and this seems to be a homework problem, I'm reluctant to also give a full proof of the second part. If you write up a decent attempt as an edit to your question, and show some real effort to try and understand the concepts involved (or at least understand what you don't understand), I can help you fill in any gaps in your attempt. – Dan Rust Oct 28 '13 at 00:57
  • I honestly would do that if I had anything to offer but the notation is confusing me and I don't understand what I need to do to prove this bit – Tom Oct 28 '13 at 01:27
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    Let's say we have a Cauchy sequence ${s^{(1)}, s^{(2)},\ldots}$ with each $s^{(j)}\in M$. Each $s^{(j)}$ is a sequence of $0$s and $1$s. We'll call the $i$-th component of $s^{(j)}$ '$s^{(j)}_i$'. Does this notation make things easier? If so I'll alter my answer to use this notation. – Dan Rust Oct 28 '13 at 01:39
  • oh I see - so I need to show that the Cauchy sequence of the sequences in M. But every Cauchy sequence converges and since this Cauchy sequence can only converge to a sequence of 1s and 0s i.e. a sequence in M, the Cauchy sequence converges in M. Is that the right idea@ – Tom Oct 28 '13 at 01:44
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    You need to show that there is a sequence which the Cauchy sequence converges to. In order to do that, you need to show that for all $N\geq 1$, for all sufficiently large $j$, all of the components with $i\leq N$ of $s_i^{(j)}$ are equal. To put it simply, you need to show that as you go down the list of sequences in the Cauchy sequence, you'll eventually find that the first $N$-string of the sequences get fixed for all $N$ (to show this, you'll need to use the Cauchy property), and so the limit of this sequence is the sequence you get by taking all the eventually fixed components. – Dan Rust Oct 28 '13 at 01:50
  • right I see. I'm not seeing how to put this together as a precise argument yet though – Tom Oct 28 '13 at 01:56
  • Feel free to upvote/accept answers which you feel have been particularly helpful. – Dan Rust Oct 28 '13 at 02:02
  • yes of course you've been ever so helpful, I understand more now but i'm still stumped by part 2 – Tom Oct 28 '13 at 02:06
  • I suggest writing out all the definitions again, write examples of Cauchy sequences of sequences (it's definitely not an easy concept to grasp when you're first introduced). Sure sometimes you get properly stuck and will never get over that hurdle, but you should try everything available before resigning. – Dan Rust Oct 28 '13 at 02:10
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Hints: i) For any $x=\{0, x_2, x_3, \ldots\} \in U_0$ onsider its $\dfrac13$-neighborhood.
ii) If $\{x^{(k)}\}$ is Cauchy in $M$, then use $|x_n^{(m)}-x_n^{(k)}| \le 2^nd(x^{(m)},x^{(k)})$.

njguliyev
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