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Let $A=\{0,1,2,...\}$ with $f$ the French railway metric that has centre $0$ and $f(a,0)=1$ for all $a\in A$ with $a\neq0$.

  1. How do I show that the metric space $(A,d)$ is complete?
  2. How do I show that $A$ is bounded, but not totally bounded?
  3. What is an example of a sequence in $A$ without convergent subsequence?
  4. What is an example of an open cover for $A$ without a finite subcover?
  5. How do I find all the compact subsets of $A$?

What I know: For the French railway metric we know that for $a,b\in A$ with $a\neq b$ we have $$f(a,b)=f(a,0)+f(0,b)=1+1=2$$ (provided that neither $a$ nor $b$ are $0$, otherwise $f(a,b)=1$).

  1. Being complete means that all Cauchy sequences in $A$ converge. I think the trick to proving this lies in the above described French metric, but I don't see it at the moment.
  2. Being totally bounded means that there are finite $a_1,...,a_n$ in $A$ such that $A=\bigcup_{i=1}^nB_\epsilon(a_i)$. It seems intuitive that this is not the case, but how do I prove this exactly?
  3. Sadly, I have no idea how to handle this one.
  4. An open cover is a collection $X$ of open subsets of $A$ such that $A\subset\bigcup_{U\in X}U$. Then we need to find one such that no finite subcolection of $X$ is an open cover for $A$.
  5. I was thinking we could take all the finite subsets of $A$; since being compact means that for all open covers of such a subset, there are finitely many elements in that open cover such that these elements are an open cover. But I am not at all sure of this, or even wheteher these would be all the compact subsets.
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  • $\begingroup$ Hint: Can you describe Cauchy and convergent sequences in this metric concretely? Also, what are open balls in this metric? $\endgroup$ – Ennar Mar 15 '16 at 21:58
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Hint: $$B_{\varepsilon}(0) = \{0\} \mathrm{\ if\ } \varepsilon \leq 1 \mathrm{\ and\ } B_\varepsilon(0) = A \mathrm{\ otherwise.}$$ Similarly for $a \not=0$: $$B_{\varepsilon}(a) = \{a\} \mathrm{\ if\ } \varepsilon \leq 1, B_\varepsilon(a) = \{0,a\} \mathrm{\ if\ } 1 < \varepsilon \leq 2 \mathrm{\ and\ } B_\varepsilon(0) = A \mathrm{\ otherwise.}$$ Now, for (1): Show that any Cauchy sequence is constant, thus convergent.

(2) Pick $\varepsilon \leq 1$.

(3) Take $x_n = n$.

(4) Similar to (2).

(5) Only the finite ones.

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  • $\begingroup$ (1) How would I do that? I know that for $(a_n)$ there is an $N$ such that for all $m,n\geq N$ we get $f(a_m,a_n)<\epsilon$. How does this lead to being constant? $\endgroup$ – user316173 Mar 16 '16 at 17:17
  • $\begingroup$ (2) I said "Being totally bounded means that there are finite $a_1,...,a_n$ in $A$ such that $A=\bigcup_{i=1}^nB_\epsilon(a_i)$.", must this apply for all $\epsilon>0$? If so, I understand the hint $\endgroup$ – user316173 Mar 16 '16 at 17:18
  • $\begingroup$ (3) In general, how would I prove that a sequence has no convergent subsequence? $\endgroup$ – user316173 Mar 16 '16 at 17:19
  • $\begingroup$ (4) What do you mean by this? Do you mean that $\bigcup_{a\in A}B_\epsilon(a)$ is the open cover for $\epsilon\leq 1$? $\endgroup$ – user316173 Mar 16 '16 at 17:21
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    $\begingroup$ For (1), choose $\varepsilon=0.5$, $if f(a_m,a_n) < 0.5$, then $a_m =a_n$ for all $m,n \geq N$ for some $N$. (2) must apply to all $\varepsilon$, so yes. If $(y_n)$ is a subsequence of $(n)$, then it's not constant, so by (1), not convergent. (4) Yes, $\{B_{0.5}(a)\ |\ a \in A\} = \{\{a\}\ |\ a\in A\}$ is an open cover with no finite subcover. (5) Obviously every finite set is compact. If $X$ is an infinite subset of $A$, then $\{B_{0.5(x)}\ |\ x \in X\}=\{ \{x\}\ |\ x \in X\}$ is an open cover with no finite subcover. $\endgroup$ – sqtrat Mar 16 '16 at 18:54
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HINT: For $(1)$, the key is to figure out which sequences are Cauchy. If $a,b\in A$, and $f(a,b)<1$, what can you say about $a$ and $b$?

The answer to the following question is the key to answering every other part of the problem:

  • If $a\in A$ and $0\le\epsilon\le 1$, what is the set $B_\epsilon(a)$?
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