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Let $X$ be a co-finite topological space. If $|X| \ge 2^{\aleph_0}=\mathfrak c$, then $X$ is contractible (https://en.wikipedia.org/wiki/Contractible_space) . Indeed, there is a bijection $f: X \times (0,1) \to X$; fix a point $a \in X$; define $H: X \times [0,1] \to X$ as $H(x,0)=x, \forall x \in X; H(x,1)=a,\forall x \in X ; H(x,t)=f(x,t), \forall (x,t)\in X \times (0,1)$. Then $H$ is continuous, hence a homotopy between $Id_X$ and the constant map on $X$ which takes everything to $a$. This shows that, under Continuum Hypothesis, any co-finite topological space $X$ with uncountable $X$, is contractible.

My question is : If every co-finite topological space with uncountable underlying set is contractible , then is it true that Continuum Hypothesis holds ? Or atleast, can we say that the fact "every co-finite topological space with uncountable underlying set is contractible" cannot be proved without Continuum Hypothesis ?

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    $\begingroup$ Interesting question; it looks like one of those cardinal characteristics of the continuum questions. See for instance Andreas Blass's answer to this question: mathoverflow.net/questions/48970/… $\endgroup$ – Todd Trimble Mar 24 '18 at 18:07
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Nice question!

I claim that this property does not necessarily imply CH. As Todd guessed in his comment, the answer is related to certain cardinal characteristics of the continuum.

Specifically, let us define the closed-partition number to be the size $\kappa$ of the smallest nontrivial partition of the unit interval $[0,1]$ into closed sets. That is, $\kappa$ is the smallest size of a set $I$, such that the unit interval admits a partition $$[0,1]=\bigsqcup_{i\in I}C_i$$ into at least two pairwise disjoint nonempty closed sets $C_i$.

It is a standard exercise to show that $\kappa$ is uncountable (see this nice explanation of Timothy Gowers); in other words, the unit interval is not a nontrivial union of countably many disjoint nonempty closed sets. A somewhat more refined observation is that $\text{cov}(\mathcal{M})\leq\kappa$, as explained in this MO answer of Andreas Blass. A further refined observation is that $\frak{d}\leq\kappa$, made by Taras Banakh in a comment on Andreas's answer. I'm not sure if this $\kappa$ already has a name or if it is provably equal to one of the well-known cardinal invariants.

Meanwhile, Arnie Miller proved in

that it is consistent with ZFC that $\kappa=\omega_1$, even while CH fails. So the unit interval can be the disjoint union of $\omega_1$ many nonempty closed sets, even when CH fails. This situation will be the key to answering your question.

Let's begin with the following observation.

Observation. If $X$ is a $T_1$ path-connected space with at least two points, then $X$ has size at least the closed-partition number $\kappa$.

Proof. If $f:[0,1]\to X$ is a path between two distinct points, then the sets $C_x=\{t\mid f(t)=x\}$ are disjoint closed sets, whose union is $[0,1]$. So $X$ must have size at least $\kappa$. $\Box$.

One can now characterize exactly which cofinite spaces are contractible.

Theorem. Suppose that $X$ is a cofinite space with at least two points. Then the following are equivalent.

  1. $X$ is contractible.
  2. $X$ is path connected.
  3. $X$ has size at least $\kappa$, the closed-partition number.

Proof. Clearly every contractible space is path connected. And we proved in the observation that every path-connected $T_1$ space (and the cofinite topology is $T_1$) has size at least $\kappa$.

What remains is to prove that every cofinite space $X$ of size at least $\kappa$ is contractible. Fix a closed particition $[0,1]=\sqcup_{i\in I} C_i$, where $I$ has size $\kappa$. Also fix distinct points $x_i\in X$ and another distinct point $a\in X$.

Define a map $H:X\times[0,1]\to X$ as follows. This is an analogue to the contraction defined in the OP under CH. Namely, let $H(x,0)=x$ and $H(x,1)=a$; and for other values of $t$, let $H(x,t)=x_i$, where $t\in C_i$.

I claim that this map is continuous and hence is a contraction of the space. To see that it is continuous, consider any open set in $X$, which is the complement of a finite set in $X$. The preimage of any point $x\in X$ not of the form $x_i$ or $a$ is just the point $(x,0)$, which is closed. The pre-image of $a$ is $X\times\{1\}\cup\{(a,0)\}$, which also is closed. And the pre-image of $x_i$ is $X\times C_i$, plus the point $(x_i,0)$, and this also is a closed set. So the preimage of any closed set in $X$ is a finite union of closed sets and hence is closed. And so the map $H$ is continuous, and therefore $X$ is contractible, as desired. $\Box$

Corollary. If ZFC is consistent, then it is consistent with ZFC that every uncountable cofinite space is contractible, yet CH fails.

Proof. Miller provides a model where $\kappa=\omega_1$, yet CH fails. In this model, every uncountable cofinite space has size at least $\kappa$, and so all of them are contractible, yet CH fails. $\Box$.

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  • $\begingroup$ Thanks ... I will take my time to read through the answer ... meanwhile, allow me to ask : Is it consistent with ZFC that a co-finite space of cardinality $\aleph_1$ is not contractible ? Btw, in the proof of your first observation , there is a typo in the definition of $C_x$ $\endgroup$ – user111524 Mar 24 '18 at 20:02
  • $\begingroup$ Thanks, I have now fixed the typo. And yes, since it is consistent with ZFC that $\kappa>\omega_1$, it follows by the theorem that in these models, $\aleph_1$ is not contractible with the cofinite topology. $\endgroup$ – Joel David Hamkins Mar 24 '18 at 20:12

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