From Countable Ordinals to a Ray
The real half-line can be assembled from successive unit intervals: the interval starting at \(n\) ends where the one starting at \(n+1\) begins. The long ray follows the same block-by-block idea, but its blocks are indexed not just by the nonnegative integers. They are indexed by all countable ordinals. A coordinate in \([0,1)\) records a position inside each block, and lexicographic order places the blocks in ordinal order.
Let \(\omega_1\) denote the first uncountable ordinal, and let \(\Omega=\{\alpha:\alpha<\omega_1\}\) be the set of countable ordinals. Thus \(0,1,2,\ldots,\omega,\omega+1,\ldots\) are in \(\Omega\); every \(\alpha\in\Omega\) has a successor \(\alpha+1\in\Omega\). The coordinate interval \([0,1)\) has its usual order and its subspace topology from \(\mathbb{R}\). We use the Cartesian product \(\Omega\times[0,1)\) as a set, but equip it with the lexicographic order and then the order topology—not the product topology.
The first coordinate identifies the block; the second gives a position within it. For example, \((4,\frac{2}{3})<(5,0)\), since the block indexed by \(4\) precedes the block indexed by \(5\). Within one block, \((4,\frac{2}{3})<(4,\frac{3}{4})\), because \(\frac{2}{3}<\frac{3}{4}\). The comparison is not coordinatewise: points in different blocks are ordered by their first coordinates, regardless of their second coordinates.
Each Block Is a Half-Open Unit Interval
For \(\alpha\in\Omega\), define the \(\alpha\)-block by \(B_\alpha=\{(\alpha,t):0\le t<1\}\). The start of this block is \((\alpha,0)\), and the start of the next block is \((\alpha+1,0)\). An interval \([p,q)\) in a linear order means the points \(x\) satisfying \(p\le x<q\).
Proof. Fix \(\alpha\in\Omega\). If \((\alpha,t)\in B_\alpha\), then \(0\le t<1\), so \((\alpha,0)\le(\alpha,t)\). Since \(\alpha<\alpha+1\), lexicographic order also gives \((\alpha,t)<(\alpha+1,0)\). Thus every point of \(B_\alpha\) belongs to the displayed half-open interval.
Conversely, let \((\beta,u)\) satisfy \((\alpha,0)\le(\beta,u)<(\alpha+1,0)\). The first inequality rules out \(\beta<\alpha\), because every point whose first coordinate is less than \(\alpha\) precedes \((\alpha,0)\). The strict second inequality rules out \(\beta\ge\alpha+1\): if \(\beta=\alpha+1\), then \((\beta,u)\ge(\alpha+1,0)\), with equality when \(u=0\); if \(\beta>\alpha+1\), then \((\beta,u)>(\alpha+1,0)\). Therefore \(\beta=\alpha\), and the point is in \(B_\alpha\). Every point \((\beta,u)\in L\) belongs to \(B_\beta\), and no point can belong to two blocks because its first coordinate is unique. Finally, if \(\alpha<\beta\), lexicographic order gives \((\alpha,t)<(\beta,u)\) for every \(t,u\in[0,1)\). This proves the interval description, partition, and block ordering. \(\square\)
The theorem makes precise how the blocks fit together. The right endpoint \((\alpha+1,0)\) is not part of \(B_\alpha\); it is the first point of \(B_{\alpha+1}\). There is no coordinate \((\alpha,1)\) in \(L\), because the second-coordinate interval is half-open. Each block, considered just as an ordered set, has the same order as \([0,1)\): the map \(t\mapsto(\alpha,t)\) preserves and reflects order.
Worked Example: Comparing Points in Finite Blocks
Compare \(x=(2,\frac{3}{4})\), \(y=(3,0)\), and \(z=(2,\frac{4}{5})\), and identify the blocks that contain them.
The first coordinates of \(x\) and \(z\) are \(2\), so compare their second coordinates. Since \(\frac{3}{4}=\frac{15}{20}\) and \(\frac{4}{5}=\frac{16}{20}\), we have \(x<z\). The point \(y\) has first coordinate \(3\), so the block-order rule gives \(z<y\), even though the second coordinate of \(y\) is \(0\). Hence \(x<z<y\).
The point \(x\) belongs to \(B_2=[(2,0),(3,0))\), and \(z\) belongs to that same block. The point \(y=(3,0)\) is the start of \(B_3\), not a point of \(B_2\).
Check: the interval theorem gives \((2,0)\le(2,\frac{3}{4})<(3,0)\) and \((2,0)\le(2,\frac{4}{5})<(3,0)\). It also gives \((3,0)\in[(3,0),(4,0))=B_3\). These interval memberships agree with the comparisons above.
What Happens at a Block Boundary?
Although a block has no last point, its points approach the start of the next block in the order topology. The following statement makes this boundary behavior explicit. Convergence here means that every open neighborhood of the proposed limit contains all terms from some point onward.
Proof. Write \(p=(\alpha+1,0)\), and let \(U\) be an open neighborhood of \(p\) in the order topology. The point \(p\) is not the least point of \(L\) and has points larger than it. By the definition of the order topology, \(U\) contains an interval \((a,b)\) with \(a<p<b\). If \(a<(\alpha,0)\), then every \(x_n\) lies above \(a\). If \(a\ge(\alpha,0)\), then \(a<p\) implies \(a=(\alpha,s)\) for some \(s\in[0,1)\): a point in block \(\alpha+1\) is either \(p\) or lies above \(p\), and every later block lies above \(p\). Since \(s<1\), choose \(N\in\mathbb{N}\) such that \(N>s/(1-s)\). For every \(n\ge N\), multiplication by \(1-s>0\) gives \(n(1-s)>s\), hence \(n/(n+1)>s\). Therefore \(x_n>a\) for all \(n\ge N\). In either case \(x_n<p<b\) for all sufficiently large \(n\), so \(x_n\in(a,b)\subseteq U\) eventually. This is convergence to \(p\). \(\square\)
Worked Example: The Boundary After the Finite Blocks
Consider the point \((\omega,0)\), where \(\omega\) is the first infinite ordinal. The finite blocks \(B_0,B_1,B_2,\ldots\) all precede \(B_\omega\). We verify that the sequence \(x_n=(n,\frac{1}{2})\) converges to \((\omega,0)\).
Let \(U\) be any open neighborhood of \((\omega,0)\). It contains an order interval \((a,b)\) with \(a<(\omega,0)<b\). Since every ordinal below \(\omega\) is finite, \(a=(m,s)\) for some finite ordinal \(m\) and \(s\in[0,1)\). Choose the positive integer \(N=m+1\). For each \(n\ge N\), \(n>m\), so lexicographic order gives \((n,\frac{1}{2})>(m,s)=a\). Also \(n<\omega\), so \((n,\frac{1}{2})<(\omega,0)<b\). Thus every term with \(n\ge N\) lies in \((a,b)\), and hence in \(U\).
Check: the first three terms are \((1,\frac{1}{2})\), \((2,\frac{1}{2})\), and \((3,\frac{1}{2})\). Each is in its indicated finite block, and each precedes \((\omega,0)\) because \(1,2,3<\omega\). The neighborhood argument works for every \(a=(m,s)\), not only for these three terms, so it verifies convergence rather than just the displayed comparisons.
This boundary example also illustrates why ordinal indices matter. The point \((\omega,0)\) follows all the finite-indexed blocks, even though there is no final finite block immediately before it. It is still a point of \(L\), because \(\omega\) is countable and therefore belongs to \(\Omega\). The proof used the order-topology neighborhoods directly; it did not assume that a sequence test characterizes convergence in arbitrary topological spaces.
The Finite Blocks and a Common Pitfall
The blocks with finite indices recover the usual nonnegative real line as an ordered set. This is a useful comparison, but it describes only the finite-indexed part of \(L\), not all of the long ray.
Worked Example: Turning Finite Blocks into Real Numbers
Let \(L_{\mathrm{fin}}=\{(n,t): n\in\{0,1,2,\ldots\},\ 0\le t<1\}\), with the order inherited from \(L\). Define \(f:L_{\mathrm{fin}}\to[0,\infty)\) by \(f(n,t)=n+t\), where \([0,\infty)\) has its usual real order.
For the point \((2,\frac{3}{4})\), the value is \(f(2,\frac{3}{4})=2+\frac{3}{4}=\frac{11}{4}\). For \((3,0)\), it is \(f(3,0)=3\). Since \(\frac{11}{4}<3\), these values preserve the already established order \((2,\frac{3}{4})<(3,0)\).
To check the full order claim, first take two points \((n,t)\) and \((n,u)\) in the same block. Then \((n,t)<(n,u)\) exactly when \(t<u\), and adding \(n\) gives \(n+t<n+u\). Next suppose \(n<m\). For finite integers, \(m\ge n+1\), so \(n+t<n+1\le m\le m+u\), using \(t<1\) and \(u\ge0\). Thus \((n,t)<(m,u)\) implies \(f(n,t)<f(m,u)\).
Finally, every real \(r\ge0\) has a unique integer part \(n\ge0\) and fractional part \(t\in[0,1)\) with \(r=n+t\). Hence \(f\) is onto, and uniqueness of this decomposition makes it one-to-one. Together, these computations prove that \(f\) is an order isomorphism from the finite-indexed blocks onto \([0,\infty)\). The map does not cover points whose first coordinate is an infinite countable ordinal.
Check: the input \((2,\frac{3}{4})\) maps to \(\frac{11}{4}\), which lies in \([2,3)\), and \((3,0)\) maps to \(3\), which begins the next interval. This agrees with the half-open block description.
Common pitfall. It is incorrect to treat \((\alpha,1)\) as the right endpoint belonging to \(B_\alpha\), or to write \((\alpha,1)=(\alpha+1,0)\). The first expression is not a point of \(L\), because \(1\notin[0,1)\); the second equality would equate two distinct ordered pairs. For a concrete check, when \(\alpha=2\), the point \((2,\frac{3}{4})\) is in \(B_2\) and lies below \((3,0)\), while \((3,0)\) is the first point of \(B_3\). The endpoint belongs to the next block; it is not an extra point attached to the previous one.
Check Your Understanding
Use the lexicographic order and the half-open block description in your answers.
- Which block contains \((\omega+2,\frac{1}{3})\), and which point starts the next block?
- Compare \((5,\frac{9}{10})\) and \((6,\frac{1}{10})\). Which coordinate determines their order?
- Explain why \((\alpha,1)\) is not a point of \(L\), and identify the first point of the block following \(B_\alpha\).
- For a fixed \(\alpha\), verify directly from the order definition that \((\alpha,\frac{1}{2})<(\alpha+1,0)\).
- Why does the map \((n,t)\mapsto n+t\) describe only the finite-indexed blocks and not all of \(L\)?