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!

  • $\begingroup$ 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

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.


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, $\kappa\le\mathfrak{c}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.