Pictured below are two subsets of the plane, each a compactification of the closed half-line with remainder a closed arc. I am really frustrated by my inability to prove that the space pictured on the right, an arc with a ray seesawing back and forth toward it, is not homeomorphic to the closed topologist's sine curve. How does one show this?
As well, in Nadler he refers to this object as the 'M-Continuum'. I have seen the left figure also referred to as the M-Continuum by another author, where there are 'half-length dips' between each dip of the typical sine curve. Are these two spaces homeomorphic?
I really have trouble figuring out how to compare these spaces. Can anyone help?
Here is a better image of this other 'M-Continuum': http://hyperspacewiki.org/index.php/M_continuum
The space $X$ pictured on the right is homeomorphic to the closed topologist's sine curve $S$.
Write $X = L \cup R$ with $L = \{ 0 \} \times [-1,1]$ and $R$ the ray approaching from both sides to $L$. Parametrize $R$ by $u : (0,1] \to \mathbb{R}^2$. Define $L'= L \times \{ 0 \} \subset \mathbb{R}^3$, $u' : (0,1] \to \mathbb{R}^3, u'(x) = (u(x),x)$ and $R'= u'((0,1])$. Then $X' = L' \cup R'$ is compact and the projection $p_{12} : \mathbb{R}^3 \to \mathbb{R}^2$ which forgets the third coordinate maps $X'$ homeomorphically onto $X$. The projection $p_{23} : \mathbb{R}^3 \to \mathbb{R}^2$ which forgets the first coordinate maps $X'$ homeomorphically onto $X'' = p_{23}(X')$. But $X''$ is easily seen to be homeomorphic to $S$.
The M-continuum $M$ is not homeomorphic to $S$.
Let $M = L \cup R_M$ with a ray $R_M$ touching the $x$-axis at the bottom of each half-length dip and let $S = L \cup R_S$ with a ray $R_S$. Both $R_M, R_S$ touch the lines $y = \pm1$ at their big amplitudes. We call $r \in R_S$ resp. $r \in R_M$ an extremal point of type $\pm1$ if it lies on the line $y = \pm1$. The points $r \in R_M$ at the bottom of the half-length dips are called extremal point of type $0$.
We shall need
Lemma. Each homeomorphism $g : L \to L$ extends to a homeomorphism $G : S \to S$.
Proof. Write $g(0,y) = (0,g'(y))$ with a homeomorphism $g' : [-1,1] \to [-1,1]$. Each closed segment $\sigma$ of $R_S$ between two subsequent extremal points is a homeomorphic copy of $[-1,1]$. In fact, let $p_i : \mathbb{R}^2 \to \mathbb{R}$ denote the projection onto the $i$-th coordinate. Then $p_1$ restricts to a homeomorphism $p_1^\sigma : \sigma \to J^\sigma = p_1(\sigma)$ and $p_2$ restricts to a homeomorphism $p_2^\sigma : \sigma \to [-1,1]$. $J^\sigma $ is the closed interval between the first coordinates of the given extremal points. We therefore get a homeomorphim $G^\sigma = (p_2^\sigma)^{-1} \circ g' \circ p_2^\sigma : \sigma \to \sigma$. The coordinate functions $G^\sigma_1 = p^\sigma_1 \circ G^\sigma : \sigma \to J_\sigma$ and $G^\sigma_2 = p^\sigma_2 \circ G^\sigma : \sigma \to [-1,1]$ are again homeomorphims. Note that $G^\sigma_2(x,y) = g'(y)$. The $G_\sigma$ and $g$ define a map $G : S \to S$ which is obvíously a bijection. It has the form $G(x,y) = (G_1(x,y),g'(y))$ where $G_1 : S \to [0,1]$ is obviously continuous (in fact, it maps $R_S$ homeomorphically onto $(0,1]$). This shows that $G$ is a homeomorphism.
Let us now assume that there exists a homeomorphism $h : M \to S$. We have $h(L) = L$ and $h(R_M) = R_S$. $g = h^{-1} : L \to L$ extends to a homeomorphism $G : S \to S$. Then $H = G \circ h : M \to S$ is a homeomorphism such that $H \mid_L = id$. Let $H_2(x,y)$ denotes the second coordinate of $H(x,y)$. By uniform continuity of $H$ we find $\delta > 0$ such that for all $y$ and $x < \delta$
$$(\ast) \phantom{x} \lvert y - H_2(x,y) \rvert \le \lvert (0,y) - H(x,y) \rvert = \lvert H(0,y) - H(x,y) \rvert < \frac{1}{2}$$ Now consider three subsequent extremal points $e_1, e_2, e_3$ of $R_M$ of type $+1,0,+1$ with first coordinates $x_1 < x_2 < x_3 < \delta$. Let $e_i' = H(e_i) = (x'_i,y'_i)$. By $(\ast)$ we have $y'_1, y'_3 \in (\frac{1}{2},1]$, $y'_2 \in (-\frac{1}{2},\frac{1}{2})$. Let $\tau$ denote the closed segment of $R_M$ between $e_1$ and $e_3$. Then $H(\tau)$ is the closed segment of $R_S$ between $e'_1$ and $e'_3$. $H(\tau)$ begins and ends in the strip $\Sigma = (0,1] \times (\frac{1}{2},1]$ but is not contained in it ($e'_2 \notin \Sigma$). This is only possible if $H(\tau)$ passes through an extremal point of type $-1$. Let $(x_0,y_0) \in\tau$ such that $H_2(x_0,y_0) = -1$. We have $x_0 < \delta$ and $y_0 \in [0,1]$. Therefore $H_2(x_0,y_0) = -1$ must have a distance $< \frac{1}{2}$ from $y_0$ which is impossible.
General methods to compare "arc-ray-spaces" with the closed topologist's sine curve $S$:
Start with the arc $L \subset \mathbb{R}^2$ and a ray $R \subset \mathbb{R}^2$ such that $L \cap R =\emptyset$ and $Y = L \cup R$ is compact. The ray is given by an embedding $u = (u_1,u_2) : (0,1] \to \mathbb{R}^2$.
Compactness of $Y$ is equivalent to "$R$ approaches $L$" in the sense that for each open neigborhood $W$ of $L$ the remainder $R \backslash W$ is compact, in other words that there exists $x \in (0,1]$ such that $u((0,x)) \subset W$.
First assume that there exists a strictly decreasing sequence $x_n \in (0,1]$ which converges to $0$ such that
(1) $u_2$ is strictly increasing on all intervals $[x_{2k+1}, x_{2k}]$ and strictly decreasing on all intervals $[x_{2k+2}, x_{2k+1}]$.
(2) $u_2(x_{2k}) \to 1$ and $u_2(x_{2k+1}) \to -1$ as $k \to \infty$.
Then $Y$ is homeomorphic to $S$.
The first step is to "normalize" $Y$ as shown above, that is to replace $Y$ by a homeomorphic $Y' = L \cup R'$ such that $R'$ is parametrized by $u' = (u'_1,u'_2)$ with $u'_1(t) = t$. The sequence $(t_n)$ is replaced by a sequence $(t'_n)$ with properties (1) and (2) for $u'_2$.
We can find a self-homeomorphism $h$ of $(0,1]$ such that $h(\nu_n) = x'_n$ where $\nu_n = \frac{2}{(2n +1) \pi}$. Then we can replace $u'$ by $u' h$ without changing $R'$.Thus w.l.o.g. assume that $x'_n = \nu_n$ (this will destroy $u'_1(x) = x$, but it is irrelevant).
Note that (1) implies $u_2(x_{n+1}) \ne u_2(x_n)$ for all $n$, and we let $J_n$ denote the closed interval bounded by these two points and $\phi_n : J_n \to [-1,1]$ the unique linear homeomorphism.
On each interval $I_n = [\nu_{n+1},\nu_n]$ there exists a homeomorphism $h_n$ such that $H_n = h_n \times \phi_n$ maps the segment of $R'$ lying over $I_n$ homeomorphically onto the corresponding segment of the closed topologist's curve. The $H_n$ and $id_L$ can be pasted to homeomorphism $Y' \to S$.
Next assume that there exist
(a) numbers $a > -1 $ and $b > 0$
(b) a decreasing sequence $x_n \in (0,1]$ which converges to $0$
such that
(1) $m_n = \min \{ u_2(x) \mid x \in [x_{2n},x_{2n-1}] \} \ge a$
(2) $M_n = \min(u_2(x_{2n}),u_2(x_{2n-1})) \ge m_n + b$.
Note that this requires $x_{2n} < x_{2n-1}$, but allows $x_{2n+1} = x_{2n}$.
Then $Y$ is not homeomorphic to $S$. This can be proved exactly as in the case of $M$.
If $Y$ and $S$ were homeomorphic, we would find a homeomorphism $H : Y \to S$ such that $H \mid_L = id$. We find $\delta > 0$ such that for all $y$ and $x < \delta$
$$(\ast) \phantom{x} \lvert y - H_2(x,y) \rvert < \min(a+1,\frac{b}{2}) .$$
Now choose $n$ such that $x_{2n-1} < \delta$. Let $e_1 = u(x_{2n}), e_2 = u(x_{2n-1})$ and $e_i' = H(e_i) = (x'_i,y'_i)$. By $(\ast)$ we have $y'_i > M_n - \frac{b}{2}$ because the $y$-coordinate of $e_i$ is $\ge M_n$. Let $\tau$ denote the closed segment of $R$ between $e_1$ and $e_2$. Then $H(\tau)$ is the closed segment of $R_S$ between $e'_1$ and $e'_2$. $H(\tau)$ begins and ends in the strip $(0,1] \times (M_n - \frac{b}{2},1]$ but is not contained in it (there exists $x \in [x_{2n},x_{2n-1}]$ such that $M_n \ge u_2(x) + b$, i.e. $u_2(x) \le M_n - b$ and $H_2(u(x)) < M_n - \frac{b}{2}$). This is only possible if $H(\tau)$ passes through an extremal point of type $-1$. Let $(x_0,y_0) \in\tau$ such that $H_2(x_0,y_0) = -1$. We have $x_0 < \delta$ and $y_0 \ge m_n \ge a$. Therefore $H_2(x_0,y_0) = -1$ must have a distance $< a+1$ from $y_0 \ge a$ which is impossible.