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Assuming the axiom of choice, is there a well-ordering of the reals such that every initial segment is closed for the usual topology? If the continuum hypothesis helps, we can also assume it.

An initial segment is a set of the form $\{x : x<y\}$ for some $y$, according to the well-ordering $<$.

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3 Answers 3

up vote 11 down vote accepted

No. There cannot be a strictly increasing $\omega_1$-sequence of closed sets in a topological space with a countable base. Their complements would be unions of open sets from the basis, and it is impossible to drop elements of a countable set uncountably many times.

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Thank you for your quick answer ! –  Denis Nov 26 '12 at 23:59
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The key property is that the reals are hereditarily separable HS. Every second countable space is HS but not the other way around. –  Ramiro de la Vega Nov 27 '12 at 0:15
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The answer is no. Assume there is such a well-order; then some downward-closed subset $D$ (which might be an initial segment, or might be all of $\mathbb{R}$) will be of order type $2^{\aleph_0}$, i.e., the least ordinal of continuum size. Being a subset of $\mathbb{R}$, $D$ contains a countable subset that is dense in $D$ (see for example here if you are in need of proof), say $x_1 \lt x_2 \lt \ldots$ according to how the elements appear in the well-order. Since there is no countable cofinal sequence in the order type $2^{\aleph_0}$, it must be that some proper initial segment contains all the $x_i$. This initial segment is both closed and dense, hence is all of $D$. Contradiction.

(Edited after Ramiro pointed out a glitch. Hopefully okay now.)

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Very nice. Separability is weaker than second countability. –  Goldstern Nov 27 '12 at 0:05
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Todd, there are well orderings of the reals with a countable cofinal sequence. It might be for instance that all the rationals form the tail of the well ordering. –  Ramiro de la Vega Nov 27 '12 at 0:10
    
you can even put a singleton at the tail ! –  Denis Nov 27 '12 at 0:50
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Yes, thanks both of you. The relevance of hereditary separability, pointed out by Ramiro under Goldstern's answer, makes itself felt here as well. –  Todd Trimble Nov 27 '12 at 1:01
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No.

If there was such a well order, let $r$ be the first real for which the initial segment $I$ bounded by $r$ is uncountable. Since $I$ is separable (being a subspace of the reals) we can fix a countable $D \subseteq I$ dense in $I$. But $I$ has order type $\omega_1$ which is a regular cardinal (usinge some choice), so there would be an $s \in I$ that bounds $D$. This is already a contradiction because then the clousure of $D$ would be contained in the initial segment determined by $s$, which is a proper subset of $I$.

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This r does not exist if we assume the continuum hypothesis. –  Denis Nov 27 '12 at 0:14
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You are right, if we assume CH this $r$ might not exist (IF the given well order happens to have type exactly $\omega_1$); in that case just ignore $r$ and take $I$ as the whole real line. –  Ramiro de la Vega Nov 27 '12 at 0:19
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