Continuum Hypothesis and the fact that every co-finite topological space, with uncountable underlying set , is contractible 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 ? 
 A: 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


*

*Miller, Arnold W., Covering
$2^\omega$ with $\omega_1$ disjoint closed
sets, The Kleene
Symp., Proc., Madison/Wis. 1978, Stud. Logic Found. Math., Vol.
101, 415-421 (1980).
ZBL0444.03026.


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.


*

*$X$ is contractible.

*$X$ is path connected.

*$X$ 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.
$\Box$.
