Let $\frak T$ be a totally ordered set of topologies on $\Bbb R$.

Is $|\frak T|\le |\Bbb R|$?

  • $\begingroup$ I suspect it can be much larger, perhaps greater than 2^R. $\endgroup$ – The Masked Avenger Aug 28 '14 at 21:32
  • $\begingroup$ Note, though, that we can bound the cofinality of $\mathfrak{I}$: pick a well-ordered cofinal sequence through $\mathfrak{I}$. Each successor topology in that sequence must contain some set of reals not in any previous topology in that sequence, and so the length of the sequence is at most $\vert 2^{\vert\mathbb{R}\vert}\vert$. $\endgroup$ – Noah Schweber Aug 28 '14 at 21:54
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    $\begingroup$ One small remark: Call a collection of sets $\mathcal{X}\subseteq\mathcal{P}(\mathbb{R})$ independent if for each $A\in\mathcal{X}$, $A$ is not in the topology generated by $\mathcal{X}-\{A\}$. Obviously any independent set yields a chain of topologies; unfortunately, there are "small" maximal independent sets - for example, $\{\{r\}: r\in\mathbb{R}\}$ - and I don't see an obvious way to build large independent sets. But maybe there is one? $\endgroup$ – Noah Schweber Aug 28 '14 at 22:00

The answer is no. As Ashutosh notes in the comments, a modification of my original answer shows in ZFC that there is a chain longer than $|\mathbb{R}|$.

Theorem. There is a linearly order chain of topologies on $\mathbb{R}$ of size larger than $\mathbb{R}$.

Proof. Let $\kappa$ be the smallest cardinal so that $2^\kappa>|\mathbb{R}|={\frak c}=2^\omega$. It follows that the tree $T=2^{<\kappa}$ has size at most continuum, since $\kappa\leq\frak{c}$ and so $2^{<\kappa}$ is a $\kappa$-sized union of sets of size at most $\frak{c}$, and hence of size at most $\frak{c}$. Thus, we may label the nodes in $2^{<\kappa}$ with reals, and get $2^\kappa$ many paths through this tree. For each path $p$, let $X_p$ be the set of reals appearing on $p$ or to the left of $p$. So we have a linearly ordered subset of $P(\mathbb{R})$ of size $2^\kappa$, which is larger than the continuum. For each $X\subset\mathbb{R}$, let $\tau_X$ be the topology with the discrete topology on $X$ and the indiscrete topology on $\mathbb{R}-X$. Since $X\subset Y\iff \tau_X\subset \tau_Y$, the chain in $P(\mathbb{R})$ translates to a chain of topologies on $\mathbb{R}$ of size $2^\kappa$, and so we have a chain larger than $|\mathbb{R}|$, as desired. QED

So in ZFC we may find a chain of topologies longer than $|\mathbb{R}|$.

Previous answer:

The answer is no, not necessarily. In particular, if the continuum hypothesis holds (or more generally, if merely $|\mathbb{R}|<2^{\omega_1}$, which is to say that Luzin's hypothesis fails), then there is a very long chain.

Theorem. There is a linearly ordered chain of topologies on $\mathbb{R}$ of size $2^{\omega_1}$. This is strictly larger than $|\mathbb{R}|$ if the continuum hypothesis holds, or more generally, if Luzin's hypothesis fails.

Proof. Consider the tree $T=2^{<\omega_1}$ consisting of countable ordinal length binary sequences. This tree has size continuum, and so we may label each node with a real. Consider the paths through this tree, ordered from left-to-right, which is the lexical order. There are $2^{\omega_1}$ many such paths. For each path, consider the set of reals appearing on the path or to the left. This gives a linearly ordered chain in $P(\mathbb{R})$ of size $2^{\omega_1}$. For each subset $X\subset\mathbb{R}$, let $\tau_X$ be the topology consisting of the discrete topology on $X$, and the indiscrete topology on $\mathbb{R}-X$. Thus, $X\subset Y\implies \tau_X\subset \tau_Y$. So our linearly order chain in $P(\mathbb{R})$ thus transforms to a chain in the set of topologies on $\mathbb{R}$, of size $2^{\omega_1}$. QED

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    $\begingroup$ Although the answer accepted, if someone could answer the question in ZFC, or prove it is independent, it would be very welcome. $\endgroup$ – Joel David Hamkins Aug 29 '14 at 2:04
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    $\begingroup$ Hi Joel, I think you can get (in ZFC) a chain of size greater than continuum in the power set of reals by considering the tree $2^{< \kappa}$ where $\kappa$ is least cardinal with $2^{\kappa} > 2^{\omega}$. $\endgroup$ – Ashutosh Aug 29 '14 at 2:40
  • $\begingroup$ That idea would work, using the same method as in my answer, provided $2^{<\kappa}\leq\frak{c}$. $\endgroup$ – Joel David Hamkins Aug 29 '14 at 2:42
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    $\begingroup$ But isn't it clear that $2^{< \kappa} = 2^{\omega}$? $\endgroup$ – Ashutosh Aug 29 '14 at 2:55
  • $\begingroup$ You are right! I have edited my answer. $\endgroup$ – Joel David Hamkins Aug 29 '14 at 3:10

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