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Edited after SamM's comment:

Consider the topological space $\mathbb{R}^\mathbb{R}$, with the usual topology. Pick a point $x \in \mathbb{R}^\mathbb{R}$ and a neighborhood $V = V_0$ of $x$. I wish to say that there is an "increasing" neighborhood chain $V_\alpha$ outside $V_0$ (that is, $V_0 \subsetneqq V_\alpha$), ordered by strict inclusion, and we can have a chain containing at least $2^c$ many elements.

I am a bit unsure how to proceed. It seems to me that constructing one neighborhood after another will only give countably many. I think a proof might be given using the special algebraic and continuum properties of $\mathbb{R}$. My question is, can a proof be given based on the fact that $\mathbb{R}^\mathbb{R}$ has cardinality $2^c$, and not using too many of the special properties that $\mathbb{R}$ has?

Also, is there a concept of "induction" that could be used to prove this statement?

Further comments: I am a bit familiar with ordinals, and can just understand what transfinite induction means. But I have never worked with it before, so a little detail would be really appreciated!

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  • $\begingroup$ By the usual topology I presume you mean the product topology? If it is possible, then one can use transfinite induction. However, one cannot take infinite intersections of preimages of open sets from $\mathbb R$, as these need not be open in the product topology, so limit ordinals will be a problem. The open sets in the product topology are "very large", which is the biggest obstruction here. $\endgroup$ – SamM Jun 2 '16 at 7:46
  • $\begingroup$ @SamM One way forward that I see is: pick a point $p_0$ inside $V_0$, choose another neighborhood $V_1$ inside $V_0$ not containing $p_0$, and proceeding so forth. But this only seems to give countably many neighborhoods. Is it possible to apply transfinite induction in this situation? Thanks! $\endgroup$ – user344037 Jun 2 '16 at 10:13
  • $\begingroup$ It isn't so obvious that one can choose a neighbourhood $V_1$. The larger problem is at limit ordinals. You will want to take intersections, but this might not give an open set. $\endgroup$ – SamM Jun 2 '16 at 10:19
  • $\begingroup$ How is this a set theory question? $\endgroup$ – Asaf Karagila Jun 2 '16 at 12:35
  • $\begingroup$ @AsafKaragila Well, the set-theory description includes transfinite hierarchies and large cardinals. I could be wrong, but it also seemed to me that people who deal with set-theory could have a high chance of knowing an answer to this. $\endgroup$ – user344037 Jun 2 '16 at 17:01
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Claim 1: Under CH, there is such a chain.

Proof: WLOG, $V_0 = \mathbb{R}^{\mathbb{R}}$. For each $A \subseteq \mathbb{R}$, let $N_A = \{y \in \mathbb{R}^{\mathbb{R}}: (\exists i \in A) (y(i) \neq x(i) + 1 )\}$. Note that each $N_A$ is an open nbd of $x$ and if $A_0 \subset A_1$, then $N_{A_0} \subset N_{A_1}$. Assume CH or just $2^{< c} = c$. Then there is a chain $\mathcal{C} \subseteq \mathbb{R}$ (under inclusion) of size $2^c$. To see this, consider for each $\eta: c \to \{0, 1\}$, the set of $\sigma:c \to \{0, 1\}$ such that $\sigma$ is eventually zero and lexicographically smaller than $\eta$.

Claim 2: It is consistent that there is no such chain.

Proof: $\mathbb{R}^{\mathbb{R}}$ has a basis of size $c$, so every chain of open sets in $\mathbb{R}^{\mathbb{R}}$ corresponds to a chain in $\mathcal{P}(\mathbb{R})$. Mitchell has constructed a model of set theory in which there is no chain of size $2^c$ in $\mathcal{P}(\mathbb{R})$ - See William Mitchell, Aronszajn trees and the independence of the transfer property, Annals of Math. Logic, Vol. 5, 1972/73, pp. 21-46

So your question is undecidable in ZFC.

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Added: The following argument is based on the assumption that the chain is well-ordered by inclusion; this was my interpretation of the OP’s ‘“increasing” neighborhood chain’. Having now seen hot_queen’s answer, I realize that the OP may have intended only that the chain be linearly ordered, in which case hot_queen’s answer is correct.


Such a chain cannot have more than $\mathfrak{c}$ elements.

Suppose that $\{V_\xi:\xi<\kappa\}$ is a family of open sets in $\Bbb R^{\Bbb R}$ such that $\bigcup_{\xi<\eta}V_\xi\subsetneqq V_\eta$ whenever $\eta<\kappa$. For each $\eta<\kappa$ fix an $x_\eta\in V_\eta\setminus\bigcup_{\xi<\eta}V_\xi$. For each $\eta<\kappa$ there are a finite $F_\eta\subseteq\Bbb R$ and and open sets $U_\eta(t)$ in $\Bbb R$ for $t\in F_\eta$ such that if

$$B_\eta=\left\{y\in\Bbb R^{\Bbb R}:y(t)\in U_\eta(t)\text{ for each }t\in F_\eta\right\}\;,\tag{1}$$

then $x\in B_\eta\subseteq V_\eta\setminus\bigcup_{\xi<\eta}V_\xi$.

$\Bbb R$ has only $\mathfrak{c}$ finite subsets and only $\mathfrak{c}$ distinct open sets, so there are at most $\mathfrak{c}$ distinct sets $B_\eta$ as in $(1)$. Thus, if $\kappa>\mathfrak{c}$ there must be $\xi<\eta<\kappa$ such that $B_\xi=B_\eta$. But then $x_\eta\in B_\xi\subseteq V_\xi$, and $x_\eta\in V_\eta\setminus V_\xi$, which is absurd. Thus, $\kappa\le\mathfrak{c}$.

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