Problem Set 03

Mar 5, 2023

Problem. 1

The required conditions follow naturally as:

  • $d’(x,x) = \sqrt{d(x,x)} = \sqrt{0} = 0$.
  • If $d(x,y) > 0$ then $d’(x,y) > 0$ as the square root of non-zero is non-zero. Otherwise $0^2 = 0$ contradicting the fact $d’(x,y) > 0$.
  • $d’(x,y) = \sqrt{d(x,y)} = \sqrt{d(y,x)} = d’(y,x)$.
  • $d’(x,y) = \sqrt{d(x,y)} \leq \sqrt{d(x,r) + d(r,y)} \leq \sqrt{d(x,r)} + \sqrt{d(r,y)} = d’(x,r) + d’(r,y)$.

For an arbitrary open-set of $d$, $\set{ y | d(x,y) < r }$ there is an equivalent open-set of $d’$, $\set{ y | d’(x,y) < \sqrt{r} }$. For an arbitrary open-set of $d’$, $\set{ y | d’(x,y) < r }$, there is an equivalent open-set of $d$, $\set{ y | d(x,y) < r^2 }$.

Note. Some good friends assisted in solving this problem.

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Problem. 2

Lemma.   For any point $p$ in $R$, There exists a smallest element in the set $\set{ q \in E | q > p }$
Assume to the contrary that no smallest element exists. Then as the set is bounded below, the infimum exists, and is a limit point. That contradicts our hypothesis of no limit points in $E$.

Corollary.   $E \cap R^+ = E^+$ has a smallest element
By the above lemma set $p = 0$.

Corollarly.   Given $x_i \in E^+$ there exists a smallest element among $E^+ \cap \set{y | y > x_i}$
By the above lemma set $p = x_i$.

Now we have a counting scheme on $E^+$. What is remaining now is to prove every element in $E$ will be hit eventually. The following lemma suffices.

Lemma.   there are finitely many elements in $E \cap [- |a| , |a|]$
Assuming the contrary for the sake of contradiction, We get infinite elements in $E \cap [ -|a|, |a| ]$. Those are present in both $E$ and $[ -|a|, |a| ]$ by definition. Since $[ -|a|, |a| ]$ is compact we know any infinite subset has a limit point (Theorem 2.41, p. 40 in baby-rudin). But then we get a limit point in E. Contradiction.

Similarly we can prove $E \cap R^- = E^-$ is countable, and hence $E$ is countable also.

Note. 1 Professor Peng Zhou hinted the solution approach

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Note. 2 Through chatting with good friends a cleaner alternative proof can be made as, "Because E has no limit points it is closed. Assume E is uncountable. Then there is an integer n such that intersection with [n,n+1] is also uncountable. This intersection is closed and bounded, thus compact. So we can take a sequence inside this intersection and it will have a convergent subsequence contradicting the assumption on limit points"

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Problem. 3

Assume for the sake of contradiction that the process does not stop after a finite number of steps. Then the sequence $x_i$ is infinite. Consider the infinite subset $\set{x_i} = S_\delta$; By hypothesis it has a limit point in $X$, Call it $p$. So for neighbourhood $N_{\delta/4}(p)$, some point $q_1 \neq p$ is in that neighbourhood. Let $r_1 = d(p, q_1)$. Consider neighbourhood $N_{r_1/2}(p)$; Clearly $q_1$ is not in it. So there is a point $q_2 \neq q_1$ in it. We have now distinct points $q_1,q_2 \in S$ such that $d(p,q_1) \leq \delta/4$ and $d(p,q_2) \leq \delta/4$. It follows $d(q_1,q_2) \leq d(q_1,p) + d(p,q_2) \leq \delta/4 + \delta/4 = \delta/2$. But the construction of sequence $x_i$ stipulates every pair of points is of distance at least $\delta$. Contradiction.

It follows by the above lemma, that for any point $p$ in $X$, the distance between it and some $x_i$ of $S$ is strictly less than $\delta$. Therefore $p$ is covered by $N_\delta{x_i}$.

Now we prove $X$ is separable. We know for each $\delta = 1/n$, The corresponding subset $S_{1/n}$ is finite. Clearly $\cup_n S_{1/n} = S$ is countably infinite. It suffices to show, For a point $p \in X - S$, it can get arbitrarily close to points of $S$. Consider arbitrary $\delta > 0$ and its corresponding neighbourhood $N_\delta(p)$.

Take $\delta’ = \delta/2$, and $n’ > 0$ such that $1/n’ < \delta’$. Consider $N_{\delta’}(p)$. There are two cases.
Case 1: A point $q \in S_{1/n’}$ is in $N_{\delta’}(p)$, Then it is also in $N_{\delta}(p)$.
Case 2: No point $q \in S_{1/n’}$ is in $N_{\delta’}(p)$. Then for any $z \in N_{\delta’}(p)$ some point $q \in S_{1/n’}$ exists such that $d(z,q) < 1/n’$. It follows $\delta = \delta/2 + \delta/2 > \delta’ + 1/n’ > d(p,z) + d(z,q) \geq d(p,q)$. In other words, $q \in N_{\delta}(p)$.

Problem. 4

Failed to solve.

Partial idea: Establish a sequence $x, f^1(x), f^2(x), f^3(x), \dots$. If I proved it is finite then I am done, as it is necessarily the case $f^k(x) = f^{k+1}(x)$. If it is infinite then a limit point of it exists as $X$ is a compact set.