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Let $I = [0,1]$ and consider $I^{\omega}$ as a topological space under the dictionary order. Are all path components of $I^\omega$ trivial? Otherwise I expect $I^{\omega}$ to be path connected since I think it, without its endpoints, is $2$-transitive. This seems odd since $I^n$ are not path connected for finite $n > 1$.

I proved (Do linear continua contain $\mathbb{R}$? Can a nontrivial connected space have only trivial path components?) that this is true for $I^{\omega_1}$ but that proof uses the fact that $\omega_1$ has uncountably many elements.

I realized by cardinal arithmetic that the set of points between any two points of $I^{\omega_1}$ has cardinality equal to that of $2^{\mathfrak c}$ and so there cannot be a path between them since the cardinalities don't agree. Closed intervals of $I^{\omega}$ do however have the right cardinality so this argument doesn't work.

Geoffrey Sangston
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  • $[0,1]\times {0}$ is just a homeomorphic copy of $[0,1]$, etc. So there are non-trivial path components. – Henno Brandsma Jun 11 '17 at 09:28
  • @Henno Could you elaborate? That appears to be a subspace of $I^2$ which is not connected even in that case. $I^2$ does in fact have nontrivial path components because ${0} \times I$ is homeomorphic to $I$ but I'm interested in $I^\omega$, where $\omega$ is the first infinite ordinal. – Geoffrey Sangston Jun 11 '17 at 13:45
  • This ordered set is not a subspace of $\ell^2$ in any obvious way. My claim is just that the set with all coordinates $>0$ equal to $0$, is a non trivial path in your ordered space. Don't you agree ? – Henno Brandsma Jun 11 '17 at 14:29
  • I see @Henno. This space is not connected either. Consider the open interval $((\frac{1}{2},-1,0,\ldots{}), (\frac{1}{2},1,0,0,\ldots{}))$ in $I^{\omega}$. Your space under the subspace topology intersects this only at $p = (\frac{1}{2},0,0,\ldots)$ so ${p}$ is an open set. The space $I^{\omega}$ is clearly Hausdorff however so your space is Hausdorff since it is a subspace of a Hausdorff space. Therefore ${p}$ is also a closed set and hence a separation. – Geoffrey Sangston Jun 11 '17 at 14:52
  • I see. You're right I think. The topology is weirder than I realised. Infinite lexicographic products were popular in the 70's, there quite a few papers on them IIRC. – Henno Brandsma Jun 11 '17 at 14:56

2 Answers2

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The fact that $I^\omega$ has no nontrivial path components can be deduced from the fact that no nontrivial subcontinuum is second countable.

Note that for every $x \in I$ the sequences $x, 0, 0, 0, ...$ and $x, 1, 1,1, ...$ are the bounds of an open interval in $I^\omega$, and these intervals are pairwise disjoint. Hence $I^\omega$ itself is not second countable.

Every subcontinuum of a linear continuum is a closed interval, and we will show that a nondegenerate interval in $I^\omega$ contains a smaller interval that is order-isomorphic to $I^\omega$.

Let $(a_n)_{n<\omega}$ and $(d_n)_{n<\omega}$ be elements of $I^\omega$ with $(a_n) < (d_n)$ and let $m = \min\{n < \omega \mid a_n < d_n \}$. Define $b_n = c_n = a_n$ for $n < m$, $b_m = c_m = (a_m + d_m)/2$ and $b_n = 0, c_n = 1$ for $n > m$. The closed interval $[(b_n), (c_n)]$ is contained in $[(a_n), (d_n)]$ and it is easy to verify that $(x_n)_{n<\omega} \mapsto (x_{n+m+1})_{n<\omega}$ defines an order-isomorphism from $[(b_n), (c_n)]$ to $I^\omega$, which must also be a homeomorphism.

It follows that $[(a_n), (d_n)]$ is not second countable and therefore cannot be a continuous image of a compact metric space.

Niels J. Diepeveen
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This paper shows that we can order embed an ordered space $X$ into a countable lexicographic product of subsets of the reals, iff $X$ only has at most countable well-ordered or reverse well-ordered subsets. The problem is that this need not be a topological embedding. It's just an injective continuous map.

Henno Brandsma
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