Does there exist a topological space $X$ such that $X^2$ and $[0,1]$ are homeomorphic?

Question

I have proved that if $X$ is not connected then $X^2$ is not connected either. So my idea was to prove that if $X$ is connected then $X^2$ blown up any point is also connected. But I don't know whether both my question and my idea are true. So could anyone give a proof or a counterexample? Thanks.

Answer 1

A product $X\times Y$ of two connected spaces $X$ and $Y$ each with at least two points cannot have a cut point. Let $(a,b)$ be any point in $X\times Y$ and we'll show that $Z = (X\times Y)-\{(a,b)\}$ is connected.

First since $X$ and $Y$ have at least two points, let $c\in X$ with $c\neq a$ and $d\in Y$ with $d\neq b$.

Now for any point $(x,y)\in Z$ we can form a connected subspace of $Z$ containing the point $(c,d)$. If $x\neq a$, then $C=(\{x\}\times Y)\cup(X\times\{d\})$ is connected since each of $\{x\}\times Y$ and $X\times\{d\}$ is connected and they share the point $(x,d)$. If $x=a$ the same $C$ won't work since it would contain $(a,b)$, but in this case $y\neq b$ so we can instead use $C=(X\times\{y\})\cup(\{c\}\times Y)$. In either case, $C$ is a connected subspace of $Z$ containing $(x,y)$ and $(c,d)$. This means that every point $(x,y)\in Z$ must be in the same connected component as $(c,d)$ so $Z$ must be connected.

This fact, together with the fact that $X\times Y$ being connected requires $X$ and $Y$ to be connected, is enough to prove that $[0,1]$ is not homeomorphic to any product other than a trivial product with a singlet $[0,1]\times\{0\}$.

Answer 2

I can think of a few proofs of the fact that there is no space $X$ with $[0,1]$ homeomorphic to $X^2$. There are probably more proofs in addition to the proofs in the other answers.

Proof 1-1: Suppose that $[0,1]\simeq X^2$. Then every connected closed subset of $[0,1]$ with more than 1 point is homeomorphic to $[0,1]$ itself. Therefore, since $X\times\{x_0\}$ and the diagonal $\{(x,x):x\in X\}$ are both connected closed subsets of $[0,1]$ with more than 1 point, these spaces are both homeomorphic to $[0,1]$. But $X\times\{x_0\},\{(x,x):x\in X\}$ homeomorphic to $X$, so $X\simeq[0,1]$ which means that $[0,1]\simeq[0,1]^2$. We know this is not the case for several reasons.

Proof 1-2: Suppose that $[0,1]\simeq X^2$. Then every set of the form $f[[0,1]]$ for some non-constant continuous $f:[0,1]\rightarrow[0,1]$ is homeomorphic to $[0,1]$. Now define $g:X^2\rightarrow X^2,h:X^2\rightarrow X^2$ by setting $g(x,y)=(x,x)$ and $h(x,y)=(x,x_0)$ for some fixed $x_0\in X$. Then $g[X^2],h[X^2]$ are both homeomorphic to $[0,1]$, but $g[X^2],h[X^2]$ are also homeomorphic to $X$. Therefore, $[0,1]\simeq[0,1]^2$ which is not the case.

Proof 2: Suppose that $X^2\simeq[0,1]$. The mapping $g:X^2\rightarrow X^2$ defined by $g(x,y)=(y,x)$ is a non-identity auto-homeomorphism with $g^2=1$ and where the set of fixed points of $g$ is homeomorphic to the diagonal $X$ which is uncountable and connected. On the other hand, if $h:[0,1]\rightarrow[0,1]$ is a non-identity auto-homeomorphism, then either $h$ is order preserving or order reversing. If $h$ is order preserving and $x<h(x)$, then $x<h(x)<h^2(x)$. Therefore, if $h$ is order preserving and $h^2=1$, then $h$ must be the identity function. Therefore, if $h$ is a non-identity auto-homeomorphism of $[0,1]$, then $h$ must be order reversing, but that means that $h$ may only have one fixed point. Therefore, since $h,g$ cannot have homeomorphic sets of fixed points, we conclude that $X^2$ cannot be homeomorphic to $[0,1]$.

Proof 3: Given two points $x,y$ in $[0,1]$, there is always a smallest closed connected subset of $[0,1]$ containing the points $x,y$. This is not the case with $X^2$. Let $x_0,x_1\in X$ be distinct. Then $\{(x,x):x\in X\},\{(x,x_0):x\in X\}\cup\{(x_1,x):x\in X\}$ are two closed connected subsets of $X^2$ containing $(x_0,x_0),(x_1,x_1)$, but the intersection of these sets is $\{(x_0,x_0),(x_1,x_1)\}$ which is neither closed nor connected.

Proof 4: There exists two partitions $P,Q$ of $X^2$ into continuumly many closed connected subsets. Furthermore, these two partitions $P,Q$ are orthogonal in the sense that if $R\in P,S\in Q$, then $R\cap S$ is a singleton. These partitions are correspond to the kernels of retractions from $X^2$ to $X\times\{x_0\},\{x_0\}\times X$. This cannot happen with $[0,1]$ for several reasons. There is no collection of uncountably many closed connected pairwise disjoint subsets of $[0,1]$ since the union of that collection of subsets would have infinite Lebesgue outer measure. There is no partition of $[0,1]$ into infinitely many closed connected sets each with more than one element.

Proof 5: There exists three closed connected subsets $C,D,E\subseteq X^2$ where $C\cap D=C\cap E=D\cap E=z$ for some $z\in X^2$. This cannot happen in $[0,1]$.

Answer 3

This is a comment but I am not entitled. If you are prepared to use (standard) machinery, then metric dimension gives a negative answer. A more elementary argument can be given using the fact the interval has precisely 2 points which are not cut points. (A cut point is one whose removal leaves a non-connected space behind).