The Ordinal Interval [0, Omega] as a Convergent Sequence

The Endpoint \(\omega\) and Its Neighborhoods

The interval \([0,\omega]\) is the simplest ordinal interval with a limit ordinal as its right endpoint. Its points are the finite ordinals \(0,1,2,\ldots\), together with \(\omega\), the first infinite ordinal. In the order topology, every finite point is isolated, while neighborhoods of \(\omega\) must contain all sufficiently large finite ordinals. This is exactly the pattern of a convergent sequence together with its limit.

We give \([0,\omega]\) its order topology, inherited from its ordinal order. Write \(\mathbb{N}_0=\{0,1,2,\ldots\}\), so that \([0,\omega]=\mathbb{N}_0\cup\{\omega\}\). The basic sets for the order topology are open intervals and, at an endpoint, the corresponding open rays, interpreted relative to \([0,\omega]\).

To see why these are the relevant basic sets, an order-topology neighborhood of the greatest point \(\omega\) contains a basic final interval \((k,\omega]\) for some \(k<\omega\). Every ordinal below \(\omega\) is finite, so \(k\in\mathbb{N}_0\). Conversely, \((k,\omega]\) is a basic open ray relative to this interval and contains \(\omega\). Thus every open neighborhood of \(\omega\) contains one of the stated tails.

Proof. Suppose first that \(U\) is open. If \(\omega\in U\), the neighborhood-basis description gives some \(k\in\mathbb{N}_0\) for which \(T_k\subseteq U\). If \(\omega\notin U\), there is no additional condition to verify. Conversely, suppose that \(\omega\notin U\). Each point \(n\in U\) is isolated: for \(n=0\), the relative order interval \([0,1)=\{0\}\) is open, and for \(n\geq1\), \((n-1,n+1)\cap[0,\omega]=\{n\}\) is open. Therefore \(U\), a union of open singletons, is open. Now suppose \(\omega\in U\) and \(T_k\subseteq U\). The point \(\omega\) has the neighborhood \(T_k\subseteq U\), and every finite point of \(U\) has an open singleton contained in \(U\). Hence each point of \(U\) has an open neighborhood contained in \(U\), so \(U\) is open. \(\square\)

Checking Neighborhoods Directly

The endpoint condition is the only restriction on an open set. A set of finite ordinals can be any subset of \(\mathbb{N}_0\), including an infinite subset with gaps of any size. If a set also contains \(\omega\), it is open precisely when it includes every finite ordinal beyond some point.

The Convergent-Sequence Space

Let \(S=\{0\}\cup\{1/j:j\in\mathbb{N}\}\subseteq\mathbb{R}\), with the subspace topology inherited from the standard topology on \(\mathbb{R}\). Thus a set is open in \(S\) exactly when it is the intersection of \(S\) with an open subset of \(\mathbb{R}\). The points \(1/j\) are isolated in \(S\), and every open neighborhood of \(0\) contains all \(1/j\) for sufficiently large \(j\).

Proof. First, \(f\) is a bijection: \(\omega\) maps to \(0\), and the finite ordinals \(n\in\mathbb{N}_0\) map to the distinct points \(1/(n+1)\), which range over all \(1/j\) for \(j\in\mathbb{N}\). We verify continuity and continuity of the inverse using neighborhoods.

Every finite ordinal \(n\) is isolated, so \(\{n\}\) is an open neighborhood of \(n\). Its image \(f(\{n\})=\{1/(n+1)\}\) is open in \(S\): for \(j=n+1\), choose a positive radius smaller than half the distance from \(1/j\) to either adjacent sequence value that exists. For \(j=1\), only the adjacent value \(1/2\) needs to be avoided. The resulting real interval intersects \(S\) only in \(\{1/j\}\). In particular, the singleton \(\{1/(n+1)\}\) is a neighborhood of \(f(n)\) for which its preimage is the open set \(\{n\}\).

It remains to check the two points corresponding to the non-isolated endpoint and limit. Let \(O\) be any open neighborhood of \(0=f(\omega)\) in \(S\). By the subspace topology, there is an \(\varepsilon>0\) such that \(S\cap(-\varepsilon,\varepsilon)\subseteq O\). Choose \(k\in\mathbb{N}_0\) large enough that \(1/(k+2)<\varepsilon\). For every \(n>k\), we have \(n+1\geq k+2\), so \(0<1/(n+1)\leq1/(k+2)<\varepsilon\). Thus \(f(T_k)\subseteq O\), and \(T_k\) is an open neighborhood of \(\omega\). This proves continuity at \(\omega\).

For continuity of the inverse at \(0\), let \(T_k\) be any basic neighborhood of \(\omega\). The image \(f(T_k)\) is \(\{0\}\cup\{1/(n+1):n>k\}\). Set \(\varepsilon=\bigl(1/(k+1)+1/(k+2)\bigr)/2\), which satisfies \(1/(k+2)<\varepsilon<1/(k+1)\). Consequently, \(S\cap(-\varepsilon,\varepsilon)=f(T_k)\): the tail values are at most \(1/(k+2)\), while all omitted sequence values are at least \(1/(k+1)\). Hence \(f(T_k)\) is open in \(S\). The map and its inverse are continuous at all finite points because those points and their images are isolated, and the endpoint checks handle the remaining points. Therefore \(f\) is a homeomorphism. \(\square\)

A Common Indexing Pitfall

A frequent error is to describe the correspondence using \(n\mapsto1/n\) for every finite ordinal \(n\). This expression is undefined at the ordinal \(0\), and it does not give a bijection from all finite ordinals to the positive reciprocal sequence. The shift \(n\mapsto1/(n+1)\) handles the zero ordinal: \(0\) maps to \(1\), \(1\) maps to \(1/2\), and in general \(n\) maps to \(1/(n+1)\). The endpoint \(\omega\), not any finite ordinal, maps to the limit \(0\).

There is also an endpoint pitfall in writing \((k,\omega)\) as a neighborhood of \(\omega\). This interval excludes \(\omega\), so it is not a neighborhood of \(\omega\) at all. The correct final interval is \((k,\omega]\), which includes the endpoint and every finite ordinal greater than \(k\).

The correspondence is useful because it turns an order-topological description into a familiar sequence-space picture without relying on a picture for the proof. The topology is determined precisely by two facts: finite points have open singletons, and every neighborhood of \(\omega\) contains a full tail. Those facts are exactly what the reciprocal sequence has at its isolated terms and its limit.