Recall that \(S_\Omega\) denotes the minimal uncountable well-ordered set. Let \(L\) denote the ordered set \(S_\Omega \times [0, 1)\) in the dictionary order, with its smallest element deleted. The set \(L\) is a classical example in topology called the long line.
Theorem. The long line is path connected and locally homeomorphic to \(\mathbb{R}\), but it cannot be imbedded in \(\mathbb{R}\).
Let \(X\) be an ordered set; let \(a < b < c\) be points of \(X\). Show that \([a, c)\) has the order type of \([0, 1)\) if and only if both \([a, b)\) and \([b, c)\) have the order type of \([0, 1)\).
Let \(X\) be an ordered set. Let \(x_0 < x_1 < \cdots\) be an increasing sequence of points of \(X\); suppose \(b = \sup\ts{x_i}\). Show that \([x_0, b)\) has the order type of \([0, 1)\) if and only if each interval \([x_i, x_{i+1})\) has the order type of \([0, 1)\).
Let \(a_0\) denote the smallest element of \(S_\Omega\). For each element \(a\) of \(S_\Omega\) different from \(a_0\), show that the interval \([a_0 \times 0, a \times 0)\) of \(S_\Omega \times [0, 1)\) has the order type of \([0, 1)\). [Hint: Proceed by transfinite induction. Either \(a\) has an immediate predecessor in \(S_\Omega\), or there is an increasing sequence \(a_i\) in \(S_\Omega\) with \(a = \sup\ts{a_i}\).]
Show that \(L\) is path connected.
Show that every point of \(L\) has a neighborhood homeomorphic with an open interval in \(\mathbb{R}\).
Show that \(L\) cannot be imbedded in \(\mathbb{R}\), or indeed in \(\mathbb{R}^n\) for any \(n\). [Hint: Any subspace of \(\mathbb{R}^n\) has a countable basis for its topology.]
Solution
Let \(a_0\) be the least element of \(S_\Omega\).
Suppose first that \([a,c)\) has order type \([0,1)\), and let \(t\in(0,1)\) be the image of \(b\). Then \([a,b)\) and \([b,c)\) correspond to \([0,t)\) and \([t,1)\), each order-isomorphic to \([0,1)\) by affine rescaling. Conversely, if both pieces have order type \([0,1)\), map them respectively onto \[ [0,1/2)\qquad\text{and}\qquad[1/2,1) \] by order isomorphisms. Their union gives an order isomorphism \([a,c)\cong[0,1)\).
If \([x_0,b)\cong[0,1)\), the images \(t_i\) of \(x_i\) increase to \(1\) because \(b=\sup x_i\). Hence each \([x_i,x_{i+1})\) corresponds to \([t_i,t_{i+1})\), which has order type \([0,1)\). Conversely, if every \([x_i,x_{i+1})\) has order type \([0,1)\), map the \(i\)th piece order-isomorphically onto \[ [1-2^{-i},\,1-2^{-(i+1)})\qquad(i\ge0). \] These maps concatenate to an order isomorphism \([x_0,b)\cong[0,1)\).
Proceed by transfinite induction on \(a\in S_\Omega-\{a_0\}\). If \(a\) has immediate predecessor \(b\), then \[ [a_0\times0,a\times0) =[a_0\times0,b\times0)\cup[b\times0,a\times0), \] and the second piece is exactly the fiber \(\{b\}\times[0,1)\), so part (a) and the induction hypothesis apply.
If \(a\) has no immediate predecessor, its section \(S_a\) is countable by minimality of \(S_\Omega\) and has no largest element. Enumerate \(S_a=\{e_0,e_1,\ldots\}\), put \(c_0=a_0\), and choose \(c_{i+1}\in S_a\) with \(c_{i+1}>\max\{c_i,e_i\}\). Then \(c_0<c_1<\cdots\) and \(\sup_i(c_i\times0)=a\times0\). By the induction hypothesis \([a_0\times0,c_{i+1}\times0)\) has order type \([0,1)\), so by part (a) each interval \[ [c_i\times0,c_{i+1}\times0) \] has order type \([0,1)\), and part (b) yields the result for \(a\).
The argument in (c), together with (a), shows more generally that every bounded interval in \(L\) is order-isomorphic to a real interval. Thus if \(p<q\) in \(L\), the closed interval \([p,q]\) is homeomorphic to \([0,1]\). Composing such a homeomorphism with the standard path \([0,1]\to[0,1]\) gives a path from \(p\) to \(q\). Hence \(L\) is path connected.
For any \(x\in L\), choose \(p<x<q\). The same bounded-interval argument shows \((p,q)\) is order-isomorphic, hence homeomorphic, to an open interval of \(\mathbb R\). Thus every point has a neighborhood homeomorphic to an open real interval.
The long line is not second countable. For each \(a\in S_\Omega\), the vertical interval \[ \{a\}\times(1/3,2/3) \] is a nonempty open subset of \(L\), and these sets are pairwise disjoint. A second-countable space can have only countably many pairwise disjoint nonempty open sets, since each contains a distinct element of a fixed countable basis. Therefore \(L\) is not second countable.
Every subspace of \(\mathbb R^n\) is second countable, because \(\mathbb R^n\) has a countable basis and intersections with the subspace form a countable basis. Hence \(L\) cannot be embedded in \(\mathbb R^n\) for any finite \(n\), in particular not in \(\mathbb R\).