Path-Connected 1-Manifold Unembeddable in R^n (3.24.12)

James Munkres Topology, chapter 3.24, exercise 12:

12. Recall $S_Ω$ denotes the minimal uncountable well-ordered set. Let $L$ denote the ordered set $S_Ω×[0,1)$ in the dictionary order with its smallest element deleted, known as the long line.

(a) Let $X$ be an ordered set, and let $a < b < c$ be elements of $X$. Show $[a,c)≅[0,1)$ as ordered sets iff $[a,b),[b,c)≅[0,1)$.
(b) Let $x_0 < x_1 < ...$ be a sequence of increasing elements of $X$, and suppose $b=\text{sup }\{x_n\}$. Show $[x_0,b)≅[0,1)$ iff $[x_i,x_{i+1})≅[0,1)$ for all $i$.
(c) Let $a_0$ denote the smallest element of $S_Ω$. For each $a∈S_Ω$ distinct from $a_0$, show $[a_0×0,a×0)$ in $S_Ω×[0,1)$ has order type $[0,1)$.
(d) Show $L$ is path connected.
(e) Show that every point in $L$ has a neighborhood homeomorphic with an open interval in $ℝ$.
(f) Show that $L$ cannot be embedded in $ℝ^n$ for any $n$.

Proof: First, note that $[0,1)≅[a,b)$ for all real $a < b$ by composing a translation with a positive scalar multiplication. As well, $$f : [0,1) → [0,∞)$$ $$f(x)=\dfrac{1}{1-x}-1$$ is bijective and order-preserving, so $[0,1)≅[0,∞)$.

(a) ($⇒$) If $[a,c)≅[0,1)$ by the order isomorphism $f$ then $[a,b)≅[0,f(b))≅[0,1)$ by the above. Similarly, $[b,c)≅[0,1)$. ($⇐$) If $[a,b)≅[0,1)$ and $[b,c)≅[0,1)$ by the isomorphisms $f_0,f_0$ respectively, then define $f : [a,c)→[0,2)$ by $f(x)=f_n(x)+n$ where $n=0$ if $x∈[a,b)$ and $n=1$ if $x∈[b,c)$. It is clear $f$ is surjective and order preserving, so that $f$ is an order isomorphism. Since $[0,2)≅[0,1)$ the claim follows.

(b) ($⇒$) As before, when $f : [x_0,b) → [0,1)$ is an isomorphism, we have $[x_i,x_{i+1})≅[f(x_i),f(x_{i+1}))≅[0,1)$. ($⇐$) For each $i∈ℕ$ let $f_i : [x_i,x_{i+1})→[0,1)$ be the isomorphism. Define $$f : [x_0,b)→[0,∞)$$ $$f(x)=f_n(x)+n$$ where $n$ is unique such that $x∈[x_n,x_{n+1})$. As before, it is clear that $f$ is surjective and order preserving, hence an isomorphism. Since $[0,1)≅[0,∞)$ the claim follows.

(c) Suppose $a∈S_Ω \setminus \{a_0\}$ is the minimal element violating the claim. Let $S=\{s∈S_Ω~|~s < a\}$. If $S$ is finite, it is clear there is a maximal element $m$ of $S$ that is the direct predecessor of $a$, so that $[a_0×0,m×0)≅[0,1)$ by minimal assumption, and $[m×0,a×0)=m×[0,1)≅[0,1)$, implying $[a_0×0,a×0)≅[0,1)$ by (a). So we may assume that $S$ is infinite and has no maximal element.

If $X$ is a countably infinite ordered set without a maximal element, define a function $g : ℕ→X$ constructed from a bijection $f : ℕ→X$ by $g(0)=f(0)$ and $g(n+1)=f(m)$ where $m∈ℕ$ is minimal such that $f(m) > g(n)$, which $m$ must exist since each element $g(n)∈X$ precedes an infinite number of elements if $X$ has no maximal element. Then $g$ shall be called a climb of $X$, and has the property that $g(0) < g(1) < ...$ is an increasing sequence, and if $X$ is embedded in $Y$ with the least upper bound property and $X$ is bounded above, then $\text{sup }X=\text{sup }\{g(n)\}_{n∈ℕ}$ (following from the fact that $g(n) ≥ f(n)$ by induction).

So let $g$ be a climb of $S$. Then it is clear that $g(0)×0 < g(1)×0 < ...$ is an increasing sequence minimally upper bounded by $a×0$, and the claim follows from (b) with an application of (a) to show each $[g(n)×0,g(n+1)×0)≅[a_0×0,g(n+1)×0)≅[0,1)$.

(d) This shall follow from part (e) when we show each two points $p,q$ of $L$ contains a neighborhood $U$ homeomorphic to $ℝ$. For if $f : ℝ→U$ is a homeomorphism, then $f : ℝ→L$ is also continuous, and we may let $r_1=f^{-1}(p)$ and $r_2=f^{-1}(q)$, and when $g : [0,1]→ℝ$ is a path from $r_1$ to $r_2$, we see $f∘g : [0,1] → L$ is a path from $p$ to $q$.

(e) As promised in (d), we shall show further that each two points $p,q∈L$ contains a neighborhood $U$ homeomorphic to $ℝ$. Assume $p < q$. Since $S_Ω$ evidently contains no maximal element, we may assume $p < q < α×0$ for some $α∈S_Ω$. Since $[a_0×0,α×0)$ in $S_Ω×[0,1)$ has order type $[0,1)$, we see $(-∞,α×0)$ in $L$ has order type $(0,1)$. Since surjective, order-preserving maps are homeomorphisms between sets endowed with the order topology, we see $(-∞,α×0)≅(0,1)$ as topological spaces. Since $(0,1)≅(-1/2,1/2)$ by translation, and $(-1/2,1/2)≅(-1,1)$ by positive scalar multiplication, it is routine to check that $f : (-1,1)→ℝ$ defined by the individual homeomorphisms on its closed halves $[0,1)≅[0,∞)$ and $(-1,0]≅(-∞,0]$ constructed similarly as above is a homeomorphism, so that $(0,1)≅ℝ$ and the claim is verified.

(f) Note that $ℝ^n$ has the countable basis $$\bigcup_{q∈ℚ^n}\bigcup_{m∈ℕ^+}B(q,1/m)$$ so that if $L$ was homeomorphic to a subspace of $ℝ^n$, it too would possess a countable basis. Also, note that for any two bases $\mathcal{A},\mathcal{B}$ of a topology, we may construct a new basis $\mathcal{C}$ consisting of only the basis elements of $\mathcal{A}$ that are contained within some element of $\mathcal{B}$. This is to show that if there is a countable basis $\mathcal{A}$ of $L$, then we may let $\mathcal{B}$ be the open intervals (without infinity) of $L$. For each $C∈\mathcal{C}$ let $C⊆(α_C×r_c,β_C×q_c)$, and for each $α∈S_Ω$ let $S_α$ denote its slice in $S_Ω$. Now when $π_1$ is projection to $S_Ω$ we note $$\bigcup_{C∈\mathcal{C}}π_1(C)⊆\bigcup_{C∈\mathcal{C}}π_1(α_C×r_c,β_C×q_c)⊆\bigcup_{C∈\mathcal{C}} S_{β_C} ⊂ S_Ω$$ since countable unions of countable sets are countable, and $S_Ω$ is uncountable. Hence not every point of the form $α×0$ is contained in a basis element, a contradiction.$~\square$