11
$\begingroup$

Do there exist examples of non-empty, infinite spaces $X$ not equipped with the discrete topology for which $X \cong X \times X$?

$\endgroup$
2

2 Answers 2

13
$\begingroup$

Elementary examples are often zero-dimensional: $\mathbb{Q}$, $\mathbb{P}$ (= the irrationals as a subset of $\mathbb{R}$) and the Cantor set $C \subset [0,1]$ all are homeomorphic to their squares (so even every finite power of itself; even countable power, for the irrationals and $C$.). A trivial example is an infinite space in the indiscrete topology.

Many infinite-dimensional spaces also obey this: $R^\omega$ in the product topology, or any higher power as well, or the Hilbert space $\ell_2$ (not really different, as $R^\omega$ is homeomorphic (as topological spaces) to $\ell_2$).

A nice one-dimensional space due to Erdős: take all points of $\ell_2$ where all coordinates are rational. Erdős showed this space is one-dimensional and it's quite clearly homeomorphic to its square: the even and odd coordinates form copies of the space itself.

$\endgroup$
9
$\begingroup$

Take for example $X=\cup_{n\ge1}\mathbb R^n$ with the topology defined by $U\subset X$ is open if and only if $U\cap\mathbb R^n $ is open for all $n$. It is then easy to check that the map $\phi:X\times X\to X$ defined by $\phi((x_i),(y_i))=(x_1,y_1,x_2,y_2,x_3,\dots)$ is a homeomorphism here an element of $X$ is represented by $(x_i)$ where $(x_i)$ is a sequence of real numbers which is nonzero only for a finite number of $i's$. (also $X=\Pi_{n\ge1}\mathbb R$ with the product topology would also work with the same map defined as above)

For a another example we have a theorem in functional analysis which says that every separable infinite dimentional Hilbert space is isomorphic to $l_2$. Since $l_2\times l_2$ is a separable Hilbert space hence by the theorem it would be isomorphic to $l_2$ (in particular it would be homemorphic)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .