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The long ray (half of the long line) is an interesting topological space. It is defined as the order topology on $\omega_1$$\times [0, 1)$ with lexicographic order. Basically, it is an uncountable number of half open intervals glued together.

My question is about not-so-long rays, where you take a countable ordinal $\alpha$, and make the space $\alpha \times [0,1)$. This is claimed to be homeomorphic to $[0,1)$. How does one define this homeomorphism? I see how this can be defined for $\omega$, $\omega^2$, and others, but I don't know how to do it in general.

This property is used to prove, for example, that every interval on the long line is homeomorphic to an interval on the real line.

Christopher King
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2 Answers2

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Well, that's actually pretty easy.

Just note that the orders are isomorphic, since they are both complete orders with dense countable subsets and no maximum. There is an isomorphism between the countable dense subsets, and it extends uniquely to the rest of the order.

More specifically, you need to pick an order embedding of $\alpha$ into $\Bbb R$, or $[0,1)$, and then just look at the intervals defined by $[j(\gamma),j(\gamma+1))$ where $j$ is the embedding.

Asaf Karagila
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  • How do you embed $\alpha$ into $\mathbb R$? – Christopher King Aug 04 '15 at 22:15
  • The proof that every countable ordinal embeds into $\mathbb{R}$ is by transfinite induction. Let $\alpha$ be the least ordinal which cannot embed into $\mathbb{R}$, and suppose $\alpha$ is countable. Then take a sequence $\beta_i$ ($i<\omega$) of ordinals $<\alpha$, whose least upper bound is $\alpha$; such a sequence exists since $\alpha$ is countable. By assumption each $\beta_i$ is embeddable into $\mathbb{R}$; by homeomorphing $\mathbb{R}$ with $(0, 1)$, we can now embed $\beta_0$ into $(0, 1)$, $\beta_1$ into $(1, 2)$, etc. This embeds $\alpha$ into $\mathbb{R}$. – Noah Schweber Aug 04 '15 at 22:41
  • @PyRulez: http://math.stackexchange.com/questions/123969/embedding-ordinals-in-mathbbq – Asaf Karagila Aug 04 '15 at 22:45
  • @Noah: Or, you know, just show every countable linear order embeds into $\Bbb Q$. – Asaf Karagila Aug 04 '15 at 22:45
  • @AsafKaragila How do you do that? – Christopher King Aug 04 '15 at 23:02
  • @PyRulez: Enumerate the linear order as $a_n$, map $a_0$ to $0$, and by induction if you mapped $a_0,\ldots,a_{n-1}$, map $a_n$ to a rational which "behave the same" with the embedding so far, as $a_n$ do with $a_0,\ldots,a_{n-1}$. Density ensures we can find such rational. – Asaf Karagila Aug 04 '15 at 23:16
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One way to embed any $\alpha$ in $\omega_1$ into $R$ is by transfinite induction. Suppose $\beta$ can be embedded by some $f$ into $[x,y]$, whenever $x,y$ are in R with $x<y$, for any $\beta<\alpha$.The case $\alpha=\phi$ (the ordinal $0$) is trivial. For the case $\alpha=\gamma+1$, let $f$ embed $\gamma$ into $[x,(x+y)/2]$ and let $f(\gamma)=y$ if $\gamma$ is a successor ordinal or is $0$,otherwise let $f(\gamma)$ be the l.u.b. of the image of $\gamma$ . For the case of a non-zero limit ordinal $\alpha$, let $\beta(n) : n=0,1,2,...$be a strictly increasing sequence of members of $\alpha$ with l.u.b. $\alpha$, and let $x(n) : n=0,1,2,...$ be a strictly increasing real sequence with l.u.b. $y$, and with $x(0)=x$. Let $f(0)$ embed $\beta(0)+1$ into $[x(0),x(1)]$ and let $f(n+1)$ embed $(\beta(n+1)+1)/(\beta(n)+1)$ into $[x(n),x(n+1)]$. This is possible because $(\beta(n+1)+1)/(\beta(n)+1)$ is order-isomorphic to an ordinal less than $\alpha$.Now let $f$ be the union of all of the $f(n)$. There is another way but I think this is long enough.

DanielWainfleet
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