In this section, we show the non-separability of \(\mathcal M_{[0,1]}\):
Theorem 2.
The metric space \((\mathcal M_{[0,1]},d_{L})\) is not separable.
Proof. It is enough to find a certain discrete subset \(\mathbb Y \subset \mathcal M_{[0,1]}\) with the continuous cardinality.
Define two subsets, flat parts
J(n,0), and pulse parts
J(n,1) in \(\mathbb {R}^{2}\):
-
Flat part: for \(n \in \mathbb {Z}_{>0}\),
$$\begin{array}{*{20}l} J(n,0)=\left[ \frac{1}{2^{n}},\frac{1}{2^{n-1}} \right]\times \{0\}, \end{array} $$
-
Pulse part: for \(n \in \mathbb {Z}_{>0}\),
$$\begin{array}{*{20}l} J(n,1)&=\left[ \frac{3}{2^{n+1}},\frac{1}{2^{n-1}} \right]\times \{0\}\\ &\cup \left\{ \left(x,\frac{3}{2^{n+1}}-x \right) : \frac{5}{2^{n+2}} \leq x \leq \frac{3}{2^{n+1}} \right\}\\ &\cup \left\{\left(x,x-\frac{1}{2^{n}}\right): \frac{1}{2^{n}} \leq x \leq \frac{5}{2^{n+2}}\right\}. \end{array} $$
See the Figure 1.
For each \(u=(u_{n})_{n \in \mathbb {Z}_{>0}} \in \{0,1\}^{\mathbb {Z}_{>0}}\), let Y
u
be a subset in \(\mathbb {R}^{2}\) as an infinite union of flat parts and pulse parts with the origin:
$$Y_{u} = \{(0,0)\} \cup \bigcup_{n=1}^{\infty} J(n,u_{n}) \subset \mathbb{R}^{2}. $$
See the Figure 2.
We equip Y
u
with the usual Euclidean distance in \(\mathbb {R}^{2}\):
$$\begin{array}{*{20}l} {}d((x_{1},x_{2}),(y_{1},y_{2}))=\left(\left(x_{1}-y_{1})^{2}+(x_{2}-y_{2}\right)^{2}\right)^{1/2}. \end{array} $$
(3)
It is easy to check that (Y
u
,d) is a compact metric space. Let \(\mathbb Y\) be the set of isometry classes of Y
u
for all \(u \in \{0,1\}^{\mathbb {Z}_{>0}}\):
$$\mathbb{Y}=\left\{Y_{u}:u \in \{0,1\}^{\mathbb{Z}_{>0}}\right\}/\text{isometry}. $$
Now we show that \(\mathbb Y \subset \mathcal M_{[0,1]}\). For \(u \in \{0,1\}^{\mathbb {Z}_{>0}}\), let f
u
be the projection from Y
u
to [ 0,1] such that x=(x
1,x
2)↦x
1. Then it is easy to see that f
u
is bi-Lipschitz continuous and, for x,y∈Y
u
,
$$\begin{array}{*{20}l} \frac{1}{\sqrt{2}}d(x,y) \leq |\,f_{u}(x)-f_{u}(y)| \leq d(x,y). \end{array} $$
(4)
Therefore the Lipschitz distance between [ 0,1] and Y
u
is bounded by
$$d_{L}\left([\!0,1],Y_{u}\right)\leq \frac{1}{2}\log 2 \quad \forall u \in \{0,1\}^{\mathbb{Z}_{>0}}. $$
Thus we have \(\mathbb Y \subset \mathcal M_{[0,1]}\).
Now we show that \(\mathbb Y\) is discrete:
Lemma 2.
Let \(Y_{u},Y_{v} \in \mathbb {Y}.\) If
$$d_{L}(Y_{u},Y_{v})<\frac{\log \left(\sqrt{2}+1\right)-\log \sqrt{5}}{2}, $$
then u=v.
Proof 3.
Let \(u=(u_{n})_{n \in \mathbb {Z}_{>0}} \in \{1,2\}^{\mathbb {Z}_{>0}}\) and \(v=(v_{n})_{n \in \mathbb {Z}_{>0}} \in \{1,2\}^{\mathbb {Z}_{>0}}\). We show that u
n
=v
n
for all \(n \in \mathbb {Z}_{>0}\). By the assumption, there exists a bi-Lipschitz function f from Y
u
to Y
v
such that
$$ |\log \text{dil}(\,f)|+|\log \text{dil}\left(\,f^{-1}\right)| <\frac{\log \left(\sqrt{2}+1\right)-\log \sqrt{5}}{2}. $$
(5)
Let us define a subset in \(\mathbb {Z}_{>0}\) as follows:
$$ P_{u}=\{n \in \mathbb{Z}_{>0}: u_{n}=1\}. $$
Without loss of generality, we may assume that P
u
is not empty. That is, Y
u
has at least one pulse. The pulse part J(n,u
n
) of Y
u
for n∈P
u
is called n-pulse of Y
u
. We note that, by the definition of the pulse parts, the peak of the n-pulse is attained at 5/2n+2 in x-axis.
It is enough for the desired result to show that P
u
=P
v
. We show that there is a bijection F:P
u
→P
v
such that F(n)=n. To show this, we have the following three steps:
-
(i)
The first step: for n∈P
u
,
$$\begin{array}{*{20}l} f_{v}\circ f \circ f^{-1}_{u}\left (\frac{5}{2^{n+2}}\right) &\in \left\{\frac{5}{2^{m+2}}: m \in \mathbb{Z}_{>0} \right\} \\ & \cup \left\{ \frac{3}{2^{m+1}}: m \in \mathbb{Z}_{>0}\right\} \\ & \cup \left\{ \frac{1}{2^{m}}:m \in \mathbb{Z}_{>0} \right\}. \end{array} $$
-
(ii)
The second step: for n∈P
u
,
$$\begin{array}{*{20}l} f_{v}\circ f \circ f^{-1}_{u}\left (\frac{5}{2^{n+2}}\right) & \notin \left\{ \frac{3}{2^{m+1}}: m \in \mathbb{Z}_{>0}\right\}\\ & \cup \left\{ {\frac{1}{2^{m}}} : {m\in} {\mathbb{Z}_{>0}} \right\}. \end{array} $$
-
(iii)
The third step: for n∈P
u
,
$$ f_{v}\circ f \circ {f^{-1}_{u}}\left (\frac{5}{2^{n+2}}\right) =\frac{5}{2^{n+2}}\quad \text{and} \quad v_{n}=1. $$
In fact, if we show the above three statements, each maximizers 5/2n+2 of n-pulses of Y
u
are mapped to the maximizers 5/2n+2 of n-pulses of Y
v
by \(f_{v}\circ f \circ f^{-1}_{u}\). This correspondence of n-pulses defines the map F:P
u
→P
v
such that F(n)=n.
The proof of the all three steps (i)-(iii) are governed by the same scheme: (A) Assume that the statements do not hold (proof by contradiction); (B) Estimate lower bounds of the dilations of f and f
−1; (C) The lower bounds obtained in (B) contradict the inequality (5).
We start to show the first step (i). Since f is homeomorphic, the maximizer 5/2n+2 of the pulse cannot be mapped to the endpoints of [0,1] by \(f_{v} \circ f \circ f^{-1}_{u}\). Assume that, for some n∈P
u
,
$$\begin{array}{*{20}l} f_{v}\circ f \circ f^{-1}_{u}\left (\frac{5}{2^{n+2}}\right)& \notin \left\{ \frac{5}{2^{m+2}} :m\in \mathbb{Z}_{>0} \right\} \\ & \cup \left\{ \frac{3}{2^{m+1}} :m \in \mathbb{Z}_{>0}\right\} \\ & \cup \left\{ \frac{1}{2^{m}} :m \in \mathbb{Z}_{>0} \right\}, \end{array} $$
and prove (i) by contradiction. By the continuity of f, there exists 0<δ<1/2n+3 such that, for any x∈[5/2n+2−δ,5/2n+2+δ], we have
$$\begin{aligned} f_{v}\circ f \circ f^{-1}_{u}\left (x\right) &\notin \left\{ \frac{5}{2^{m+2}} :m \in \mathbb{Z}_{>0} \right\} \\ & \cup \left\{ \frac{3}{2^{m+1}}:m\in \mathbb{Z}_{>0}\right\} \\ & \cup \left\{ \frac{1}{2^{m}} :m \in \mathbb{Z}_{>0} \right\}. \end{aligned} $$
Therefore we have
$$ \begin{aligned} & d \left(f\circ f^{-1}_{u}\left(\frac{5}{2^{n+2}}-\delta\right),f\circ f^{-1}_{u}\left(\frac{5}{2^{n+2}}+\delta\right) \right)\\ &=d \left(f\circ f^{-1}_{u}\left(\frac{5}{2^{n+2}}-\delta\right),f\circ f^{-1}_{u}\left(\frac{5}{2^{n+2}}\right) \right)\\ &\quad+d \left(f\circ f^{-1}_{u}\left(\frac{5}{2^{n+2}}\right),f\circ f^{-1}_{u}\left(\frac{5}{2^{n+2}}+\delta\right) \right). \end{aligned} $$
Here we use the fact that the three points \(f\circ f^{-1}_{u}(5/2^{n+2}-\delta)\), \(f\circ f^{-1}_{u}(5/2^{n+2})\) and \(f\circ f^{-1}_{u}(5/2^{n+2}+\delta)\) are on the same line. By using the inequality (4), the dilation of f is estimated as follows:
$$\begin{aligned} \text{dil}(f) &\geq \frac{d \left(f\circ f^{-1}_{u}\left(\frac{5}{2^{n+2}}-\delta\right),f\circ f^{-1}_{u}\left(\frac{5}{2^{n+2}}+\delta\right) \right)}{d\left(f^{-1}_{u}\left(\frac{5}{2^{n+2}}-\delta\right), f^{-1}_{u}\left(\frac{5}{2^{n+2}}+\delta\right) \right)}\\ & = \frac{d \left(f\circ f^{-1}_{u}\left(\frac{5}{2^{n+2}}-\delta\right),f\circ f^{-1}_{u}\left(\frac{5}{2^{n+2}}\right) \right)}{2\delta} \\ &\quad +\frac{d \left(f\circ f^{-1}_{u}\left(\frac{5}{2^{n+2}}\right),f\circ f^{-1}_{u}\left(\frac{5}{2^{n+2}}+\delta\right) \right)}{2\delta} \\ &\geq \frac{d \left(f^{-1}_{u}\left(\frac{5}{2^{n+2}}-\delta\right), f^{-1}_{u}\left(\frac{5}{2^{n+2}}\right) \right) }{2\delta \text{dil}(f^{-1})} \\ & \quad + \frac{d \left(f^{-1}_{u}\left(\frac{5}{2^{n+2}}\right), f^{-1}_{u}\left(\frac{5}{2^{n+2}}+\delta\right) \right) }{2\delta \text{dil}(f^{-1})}\\ &= \frac{\sqrt{2}}{\text{dil}(f^{-1})}. \end{aligned} $$
In the above last line, we just calculated the distance following the Euclidean distance (3) in the n-pulse J(n,1). This implies that \(\text {dil}(\,f)\geq 2^{\frac {1}{4}}\), or \(\text {dil}(f^{-1}) \geq 2^{\frac {1}{4}}\). Thus we have
$$d_{L}(Y_{u},Y_{v}) \geq \frac{\log 2}{4}. $$
This contradicts the inequality (5). Therefore we have, for any n∈P
u
,
$$\begin{array}{*{20}l} f_{v}\circ f \circ f^{-1}_{u}\left (\frac{5}{2^{n+2}}\right) & \in \left\{ \frac{5}{2^{m+2}}: m \in \mathbb{Z}_{>0} \right\} \\ &\cup \left\{ \frac{3}{2^{m+1}} :m \in \mathbb{Z}_{>0}\right\} \\ &\cup \left\{ \frac{1}{2^{m}} :m \in \mathbb{Z}_{>0} \right\}. \end{array} $$
We start to show the second step (ii) by contradiction. Assume that, for some n∈P
u
,
$$f_{v}\circ f \circ f^{-1}_{u}\left (\frac{5}{2^{n+2}}\right) \in \left\{ \frac{1}{2^{m}} :m \in \mathbb{Z}_{>0} \right\}. $$
Then there exists \(n_{1}\in \mathbb {Z}_{>0}\) such that
$$ \begin{aligned} f_{v}\circ f \circ f^{-1}_{u}\left (\frac{5}{2^{n+2}}\right) = \frac{1}{2^{n_{1}}}. \end{aligned} $$
(6)
By the same argument as the first step (i), we can obtain \(v_{n_{1}}=1\), that is, the n
1-pulse exists in Y
v
. By the continuity of f, there exists \(0< \delta < 1/2^{n_{1}+3}\phantom {\dot {i}\!}\) such that
$${ \begin{aligned} {}f_{u} \circ f^{-1} \circ f^{-1}_{v} \!\left(\left[\frac{1}{2^{n_{1}}}-\delta,\frac{1}{2^{n_{1}}} \!\right) \!\right) &\subset \left(\frac{5}{2^{n+2}}-\frac{1}{2^{n+3}},\frac{5}{2^{n+2}} \right) \\ & \cup \left(\frac{5}{2^{n+2}},\frac{5}{2^{n+2}}+\frac{1}{2^{n+3}} \right)\!, \end{aligned}} $$
and
$${\begin{aligned} {}f_{u} \circ f^{-1} \circ f^{-1}_{v} \left(\!\left(\frac{1}{2^{n_{1}}},\frac{1}{2^{n_{1}}} +\delta \right] \right) & \subset \left(\frac{5}{2^{n+2}}-\frac{1}{2^{n+3}},\frac{5}{2^{n+2}} \right) \\ & \cup \left(\frac{5}{2^{n+2}},\frac{5}{2^{n+2}}+\frac{1}{2^{n+3}} \right)\!. \end{aligned}} $$
Noting the definition of (Y
v
,d), for \( x \in \left [\frac {1}{2^{n_{1}}}-\delta,\frac {1}{2^{n_{1}}} \right)\) and \(y \in \left (\frac {1}{2^{n_{1}}},\frac {1}{2^{n_{1}}} +\delta \right ]\), we have
$$\begin{aligned} d\left(\,f^{-1}_{v}(x),f^{-1}_{v}(y)\right)&=\left(|x-y|^{2}+\!\left|\,y-\!\frac{1}{2^{n_{1}}}\right|^{2}\right)^{\!\!1/2},\\ d\left(f^{-1}_{v}(x),f^{-1}_{v}\left(\frac{1}{2^{n_{1}}}\right)\right)&=\left|x-\frac{1}{2^{n_{1}}}\right|,\\ d\left(f^{-1}_{v}(y),f^{-1}_{v}\left(\frac{1}{2^{n_{1}}}\right)\right)&=\sqrt{2}\left|y-\frac{1}{2^{n_{1}}}\right|. \end{aligned} $$
Since \(|x-y|=|x-2^{-n_{1}}|+ |y-2^{-n_{1}}|\phantom {\dot {i}\!}\), we have the following inequality:
$$ \begin{aligned} & d\left(f^{-1}_{v}(x),f^{-1}_{v}\left(\frac{1}{2^{n_{1}}}\right)\right)+d\left(f^{-1}_{v}(y),f^{-1}_{v}\left(\frac{1}{2^{n_{1}}}\right)\right) \\ & = \left(|x-2^{-n_{1}}|^{2} + 2\sqrt{2} | x-2^{-n_{1}} || y-2^{-n_{1}} | \right. \\ &\quad\left. + \;2| y-2^{-n_{1}} |^{2} \right)^{1/2} \\ & \leq \left(\sqrt{2}|x-2^{-n_{1}}|^{2} + 2\sqrt{2} | x-2^{-n_{1}} || y-2^{-n_{1}} | \right.\\ &\quad\left. + \;2\sqrt{2} | y-2^{-n_{1}} |^{2} \right)^{1/2} \\ & = 2^{\frac{1}{4}} d\left(\,f^{-1}_{v}(x),f^{-1}_{v}(y)\right). \end{aligned} $$
(7)
On the other hand, there exist \( x_{0} \in \left [\frac {1}{2^{n_{1}}}-\delta,\frac {1}{2^{n_{1}}} \right)\) and \(y_{0} \in \left (\frac {1}{2^{n_{1}}},\frac {1}{2^{n_{1}}} +\delta \right ]\) such that
$$\begin{aligned} & d\left(f^{-1} \circ f^{-1}_{v}(x_{0}),f^{-1}_{u}\left(\frac{5}{2^{n+2}}\right) \right) \\ &= d\left(f^{-1} \circ f^{-1}_{v}(y_{0}),f^{-1}_{u}\left(\frac{5}{2^{n+2}}\right) \right). \end{aligned} $$
Thus the triangle determined by the three vertices \(f\circ f^{-1}_{v}(x_{0})\), \(f\circ f^{-1}_{v}(y_{0})\) and \(f^{-1}_{u}\left (5/2^{n+2}\right)\) is an isosceles right triangle, and we can calculate
$$\begin{array}{*{20}l} & d\left(\,f^{-1} \circ f^{-1}_{v}(x_{0}),f^{-1} \circ f^{-1}_{v}(y_{0}) \right)\\ \notag &=\frac{1}{\sqrt{2}}d\left(f^{-1} \circ f^{-1}_{v}(x_{0}),f^{-1}_{u}\left(\frac{5}{2^{n+2}}\right) \right) \\ & \quad + \frac{1}{\sqrt{2}}d\left(f^{-1} \circ f^{-1}_{v}(y_{0}),f^{-1}_{u}\left(\frac{5}{2^{n+2}} \right)\right). \end{array} $$
(8)
By (6), (7) and (8), we have a bound for the dilation of f:
$$ \begin{array}{l}\begin{array}{cc}\frac{1}{\mathrm{dil}\left(\kern0.3em f\right)}& \le \frac{d\left(\kern0.3em {f}^{-1}\circ {f}_v^{-1}\left({x}_0\right),{f}^{-1}\circ {f}_v^{-1}\left({y}_0\right)\right)}{d\left({f}_v^{-1}\left({x}_0\right),{f}_v^{-1}\left({y}_0\right)\right)}\\ {}=\frac{d\left({f}^{-1}\circ {f}_v^{-1}\left({x}_0\right),{f}_u^{-1}\left(\frac{5}{2^{n+2}}\right)\right)}{\sqrt{2}d\left(\kern0.3em {f}_v^{-1}\left({x}_0\right),{f}_v^{-1}\left({y}_0\right)\right)}\\ {}\kern1em +\frac{d\left({f}^{-1}\circ {f}_v^{-1}\left({y}_0\right),{f}_u^{-1}\left(\frac{5}{2^{n+2}}\right)\right)}{\sqrt{2}d\left({f}_v^{-1}\left({x}_0\right),{f}_v^{-1}\left({y}_0\right)\right)}\\ {}\le \frac{\mathrm{dil}\left(\kern0.3em {f}^{-1}\right)d\left(\kern0.3em {f}_v^{-1}\left({x}_0\right),{f}_v^{-1}\left(\frac{1}{2^{n_1}}\right)\right)}{\sqrt{2}d\left(\kern0.3em {f}_v^{-1}\left({x}_0\right),{f}_v^{-1}\left({y}_0\right)\right)}\\ {}\kern1em +\frac{\mathrm{dil}\left(\kern0.3em {f}^{-1}\right)d\left(\kern0.3em {f}_v^{-1}\left({y}_0\right),{f}_v^{-1}\left(\frac{1}{2^{n_1}}\right)\right)}{\sqrt{2}d\left(\kern0.3em {f}_v^{-1}\left({x}_0\right),{f}_v^{-1}\left({y}_0\right)\right)}\\ {}\le \frac{1}{2^{\frac{1}{4}}}\mathrm{d}\mathrm{i}\mathrm{l}\left(\kern0.3em {f}^{-1}\right).\end{array}\end{array} $$
(9)
Here we used the equality (8) in the second and third lines, the equality (6) and the definition of the dilation in the fourth and fifth lines, and the inequality (7) in the last line. The inequality (9) implies that \(\text {dil}(f)\geq 2^{\frac {1}{8}}\) or \(\text {dil}(f^{-1}) \geq 2^{\frac {1}{8}}\). Thus we have
$$d_{L}(Y_{u},Y_{v}) \geq \frac{\log 2}{8}. $$
This contradicts the inequality (5). Therefore we have, for any n∈P
u
,
$$f_{v}\circ f \circ f^{-1}_{u}\left (\frac{5}{2^{n+2}}\right) \notin \left\{ \frac{1}{2^{m}} :m \in \mathbb{Z}_{>0} \right\}. $$
By the same argument as above, we have, for any n∈P
u
,
$$f_{v}\circ f \circ f^{-1}_{u}\left (\frac{5}{2^{n+2}}\right) \notin \left\{ \frac{3}{2^{m+1}} :m \in \mathbb{Z}_{>0}\right\}. $$
Now we start to show the third step (iii). By the above two steps (i) and (ii), we have that, for any n∈P
u
, there exists \(p_{f}(n) \in \mathbb {Z}_{>0}\) such that \( f_{v}\circ f \circ f^{-1}_{u}\left (\frac {5}{2^{n+2}}\right) =\frac {5}{2^{p_{f}(n)+2}}.\phantom {\dot {i}\!} \)
By the same argument as the first step (i), we can check that p
f
(n)∈P
v
, that is, \(v_{p_{f}(n)}=1\). Also for the inverse function f
−1, we have that, for any n∈P
v
, there exists \(p_{f^{-1}}(n) \in P_{u}\phantom {\dot {i}\!}\) such that
$$f_{u} \circ f^{-1} \circ f^{-1}_{v}\left (\frac{5}{2^{n+2}}\right) =\frac{5}{2^{p_{f^{-1}}(n)+2}}. $$
Since f is a bijection, the map p
f
is a bijection from P
u
to P
v
and \(p^{-1}_{f}=p_{f^{-1}}\).
Now it suffices to show that p
f
(n)=n for all n∈P
u
. We assume that there exists l∈P
u
such that p
f
(l)≠l. Without loss of generality, we may assume p
f
(l)>l. We first show that
$$\begin{array}{*{20}l} f_{u} \circ f^{-1} \circ f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}\right) \in \left(\frac{1}{2^{l}},\frac{5}{2^{l+2}}\right) \cup \left(\frac{5}{2^{l+2}},\frac{3}{2^{l+1}}\right)\!. \end{array} $$
(10)
To show this, it suffices to show that
$$\frac{\sqrt{2}}{2^{l+2}} > d\left(f^{-1} \circ f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}\right),f^{-1}_{u}\left(\frac{5}{2^{l+2}}\right)\right), $$
where the above inequality means that the point \(f^{-1} \circ f^{-1}_{v}\left (\frac {1}{2^{p_{f}(l)}}\right)\) belongs to one of two edges in the l-pulse crossing at the right angle. By \(p_{f^{-1}}\circ p_{f}(l)=p^{-1}_{f}\circ p_{f}(l)=l\), we have
$${\begin{aligned} {}\frac{\sqrt{2}\text{dil}\left(f^{-1}\right)}{2^{p_{f}(l)+2}} &= \text{dil}\left(f^{-1}\right)d\left(f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}\right),f^{-1}_{v}\left(\frac{5}{2^{p_{f}(l)+2}}\right)\right)\\ &\ge d\left(f^{-1} \circ f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}\right),f^{-1}\circ f^{-1}_{v}\left(\frac{5}{2^{p_{f}(l)+2}}\right)\right) \\ &= d\left(f^{-1} \circ f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}\right),f^{-1}_{u}\left(\frac{5}{2^{p_{f^{-1}}\circ p_{f}(l)+2}}\right)\right). \\ &= d\left(f^{-1} \circ f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}\right),f^{-1}_{u}\left(\frac{5}{2^{l+2}}\right)\right). \end{aligned}} $$
Since we have \(\text {dil}\left (f^{-1}\right)\leq 2^{\frac {1}{4}}\) (by the inequality (5)) and p
f
(l)≥l+1, it holds
$${}\frac{\sqrt{2}}{2^{l+2}} \! >\! \frac{\sqrt{2}\text{dil}(f^{-1})}{2^{p_{f}(l)+2}} \!\ge\! d\left(f^{-1} \!\circ f^{-1}_{v}\!\left(\frac{1}{2^{p_{f}(l)}}\right)\!,f^{-1}_{u}\left(\frac{5}{2^{l+2}}\right)\right). $$
Thus we have shown (10).
By the continuity of f and (10), there exists δ>0 such that \(\delta < \frac {1}{2^{p_{f}(l)+3}}\) and
$$\begin{array}{*{20}l} & f_{u} \circ f^{-1} \circ f^{-1}_{v}\left(\left[\frac{1}{2^{p_{f}(l)}}-\delta,\frac{1}{2^{p_{f}(l)}}+\delta\right]\right) \\ & \subset \left(\frac{1}{2^{l}},\frac{5}{2^{l+2}}\right) \cup \left(\frac{5}{2^{l+2}},\frac{3}{2^{l+1}}\right). \end{array} $$
Since the three points \(f^{-1} \circ f^{-1}_{v}\left (\frac {1}{2^{p_{f}(l)}}-\delta \right), f^{-1} \circ f^{-1}_{v}\left (\frac {1}{2^{p_{f}(l)}} \right)\) and \( f^{-1} \circ f^{-1}_{v}\left (\frac {1}{2^{p_{f}(l)}}+\delta \right)\) are on the same line, we have
$$ \begin{aligned} &d\left(f^{-1} \circ f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}-\delta \right),f^{-1} \circ f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}+\delta \right)\right)\\ &= d\left(f^{-1} \circ f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}-\delta \right), f^{-1} \circ f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}} \right)\right)\\ &\quad+d\left(f^{-1} \circ f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}\right),f^{-1} \circ f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}+\delta \right)\right)\!. \end{aligned} $$
(11)
Thus the inclusion (10) and the equality (11) imply the following bound of the dilation of f:
$$\begin{aligned} &\text{dil}(f^{-1})\\ &\geq \frac{d \left(f^{-1} \circ f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}-\delta\right),f^{-1} \circ f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}+\delta\right) \right)}{d\left(f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}-\delta\right), f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}+\delta\right) \right)}\\ & = \frac{d \left(f^{-1} \circ f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}-\delta\right),f^{-1} \circ f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}\right) \right) }{\sqrt{5}\delta}\\ &\quad +\frac{d \left(f^{-1} \circ f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}\right),f^{-1} \circ f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}+\delta\right) \right)}{\sqrt{5}\delta}\\ &\geq \frac{d \left(f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}-\delta\right), f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}\right) \right) }{\sqrt{5}\delta \text{dil}(f)} \\ & \quad + \frac{d \left(f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}\right), f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}+\delta\right) \right) }{\sqrt{5}\delta \text{dil}(f)}\\ & = \frac{\sqrt{2}+1}{\sqrt{5}\text{dil}(f)}. \end{aligned} $$
Here we used the following equality in the third and fourth lines:
$$\begin{aligned} & d\left(f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}-\delta\right), f^{-1}_{v}\left(\frac{1}{2^{p_{f}(l)}}+\delta\right)\right) \\ &=\left(\left|\frac{1}{2^{p_{f}(l)}}-\delta-\left(\frac{1}{2^{p_{f}(l)}}+\delta\right)\right|^{2}+|\delta|^{2}\right)^{1/2}\\ &=\sqrt{5}\delta. \end{aligned} $$
Thus \(\text {dil}(f) \geq \left (\frac {\sqrt {2}+1}{\sqrt {5}}\right)^{1/2}\) or \(\text {dil}(f^{-1}) \geq \left (\frac {\sqrt {2}+1}{\sqrt {5}}\right)^{1/2}\).
This contradicts the inequality (5). Therefore we have p
f
(n)=n for any n∈P
u
.
We have completed all of the three steps. Setting F(n)=p
f
(n), we have that the map F:P
u
→P
v
is a bijection such that F(n)=n and this implies P
u
=P
v
. We have completed the proof.
We resume the proof of Theorem 2.
Proof of Theorem 2.
By using Lemma 2, we know that \((\mathbb Y, d_{L})\) is discrete. Since the cardinality of \(\mathbb Y\) is continuum and \(\mathbb Y \subset \mathcal M_{[0,1]}\), we have that \((\mathcal M_{[0,1]}, d_{L})\) is not separable. We have completed the proof. □
Remark 1.
Theorem 2 says that \(\mathcal M_{[0,1]}=\{X \in \mathcal M: d([0,1],X)<\infty \}\) is not separable. By the proof of Theorem 2, moreover we know the following stronger result:
Let \(B_{d_{L}}([0,1], \delta)\) denote the ball in \(\mathcal M_{[0,1]}\) centered at [0,1] with radius δ>0 with respect to the Lipschitz distance d
L
, that is,
$$B_{d_{L}}([0,1], \delta)=\{X \in \mathcal M_{[0,1]}: d_{L}([0,1], X)<\delta\}. $$
Then, for any δ>0, B([ 0,1],δ) is not separable.
In fact, let
$$ \begin{aligned} J^{\epsilon}(n, 1)=&\left[3/2^{n+1}, 1/2^{n-1}\right] \times \{ 0 \} \\ &\cup \left\{ (x, \epsilon (3/2^{n+1}\,-\,x): 5/2^{n+2}\le x \le 3/2^{n+1} \right\} \\ & \cup \left\{ (x, \epsilon (x-1/2^{n})): 1/2^{n} \le x \le 5/2^{n+2} \right\}, \\ J^{\epsilon}(n,0)=&\;J(n,0), \\ Y^{\epsilon}_{u}\!=&\;\{(0, 0)\} \cup \bigcup_{n=1}^{\infty}\!J^{\epsilon}(n, u_{n}),\,\, u\,=\,(u_{n})_{\!n \in \mathbb{Z}_{>0}} \!\!\in\!\! \{0,1\!\}^{\mathbb{\!Z}_{>0}}\!. \end{aligned} $$
Then, by the similar proof to that of Theorem 2, we obtain (i) For every ε>0, the set
$$\mathbb Y^{\epsilon}=\left\{Y^{\epsilon}_{u}: u \in \{0, 1\}^{\mathbb{Z}_{>0}}\right\} /\text{isometry} $$
is discrete with cardinality of the continuum. (ii) For every δ>0, there exists ε>0 such that \(\mathbb Y^{\epsilon } \subset B_{d_{L}}([0,1], \delta)\).
The statement (ii) implies that \(B_{d_{L}}([0,1], \delta)\) is not separable for any δ>0.