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I noticed that the natural order of the Reals alone, being complete ( satisfying LUB ) , is able to prove that the induced order topology is complete ( every Cauchy sequence converges ). We are able to talk about completeness of such topology, because it ends up being uniformizable ( even more, it is metrizable ).

Now, i'm asking myself if any total order that is complete can induce a topology that is at least uniformizable, so that it will be able to prove the completeness of its induced uniform structure.

Finally, i'm a real beginner in topology and real analysis, i don't even know if there could be any possible example of a complete order whose induced order topology ( even if uniformizable ) that would be interesting ... I'm crawling in the dark but at the same time just trying to see to what degree of generality things hold.

Thanks in advance.

Ludolila
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nerdy
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Let $\langle X,\le\rangle$ be a linear order, and let $\tau$ be the order topology on $X$ induced by $\le$: $\tau$ has as a subbase $\{(\leftarrow,x):x\in X\}\cup\{(x,\to):x\in X\}$. Then the space $\langle X,\tau\rangle$ is $T_5$, i.e., $T_1$ and hereditarily normal; you can find proofs here. In particular, $X$ is completely regular and therefore uniformizable. None of this requires that the order be complete.

If the order is complete, including endpoints, then $X$ is compact; if it’s complete in the weaker sense that every set bounded above has a supremum, then it need not be compact, since it need not have endpoints, but it satisfies a Heine-Borel theorem: a subset of $X$ is compact if and only if it’s closed and bounded. In particular, $X$ is locally compact, so it satisfies the Baire category theorem, just as if it were a complete metric space.

Community
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Brian M. Scott
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  • Thanks for that. Do you know any example of another uniform structure ( metric space or not ) , unrelated to the reals , that is complete and induced by some natural order or perhaps the reals is the only example? Is it more of a characteristic of the reals and i shouldn't be looking elsewhere ? Thanks a lot in advance – nerdy Apr 04 '15 at 17:58
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    @nerdy: Take any compact linearly ordered space: since it’s compact and Hausdorff, it has a unique uniformity, which by compactness is necessarily complete. There are many compact linearly ordered spaces that have nothing to do with the reals, e.g., $\omega_1+1$ in its natural order, where $\omega_1$ is the first uncountable ordinal. – Brian M. Scott Apr 04 '15 at 18:03
  • Just to confirm. So every linearly ordered set that is dedekind complete ( the weaker kind of completeness ) induces an order topology is at least locally compact , correct ? – nerdy Apr 05 '15 at 01:52
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    @nerdy: Yes, that’s correct. – Brian M. Scott Apr 05 '15 at 16:29