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 \(d\), \(\{ y \mid d(x,y) < r \}\) there is an equivalent open-set of \(d'\), \(\{ y \mid d'(x,y) < \sqrt{r} \}\). For an arbitrary open-set of \(d'\), \(\{ y \mid d'(x,y) < r \}\), there is an equivalent open-set of \(d\), \(\{ y \mid d(x,y) < r^2 \}\).
Note. Some good friends assisted in solving this problem.
Problem 2
Lemma. For any point \(p\) in \(R\), There exists a smallest element in the set \(\{ q \in E \mid q > p \}\)
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 \(E^+ \cap \{y \mid y > x_i\}\)
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
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".
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 \(\{x_i\} = S_\delta\); 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
Proposition 1. The distance function \(d: X \times X \rightarrow R\) in a metric space \(X\) is continuous.
Proof. Fix \((a,b)\). Let \(\epsilon > 0\). We can take small enough \(\delta\) such that \(d(a,x) < \epsilon/2\) and \(d(b,y) < \epsilon/2\). By the triangular inequality \(d(x,y) \leq d(x,a) + d(a,b) + d(b,y)\). Hence \(| d(x,y) - d(a,b) | < | d(x,a) + d(b,y) | < \epsilon\).
Proposition 2. The function \(g(x) = d(x,f(x))\) is continuous over \(X\).
Proof. Define a vector-valued function \(h(x) = ( h_1(x), h_2(x) )\) where \(h_1(x) = x\) is the identity and \(h_2(x) = f(x)\). Then \(h\) is continuous, and so is the composite function \(g = d \circ h\).
Theorem. Problem statement.
Proof. As before let \(g: X \rightarrow \mathbb{R}\) by \(x \mapsto d(x, f(x))\). The image \(\{ d(x, f(x)) \mid x \in X \}\) is lower-bounded by \(0\). Since \(X\) is non-empty and \(\mathbb{R}\) has the greatest-lower-bound property, It follows \(inf \, X = m\) exists. Assume for contradiction \(m > 0\). By thm 4.16, p 89, rudin, there is a point \(p_0 \in X\) where \(g(p_0) = m = d(p_0, f(p_0))\). But we are given \(d(f(p_0), f^2(p_0)) < m\), i.e \(g(f(p_0)) < g(p_0)\). Contradiction. Therefore \(inf \, X = 0\) and there’s a point \(p\) such that \(g(p) = d(p, f(p)) = 0\), implying \(f(p) = p\).