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A few days ago, I was trying to generalize the defintion of Euclidean spaces by trying to define $\mathbb{R}^{0.5}$.

Question: Is there a metric space $A$ such that $A\times A$ is homeomorphic to $\mathbb{R}$?

I am interested also in seeing examples of $A$ which are only topological spaces

Edit: If there exists a topological space $A$ such that $A\times A\cong \Bbb R$, then $A\times \{a\}$ is a subspace of $A\times A$ ($a\in A$). Hence $A\times\{a\}$ can be embedded in $\mathbb{R}$, since $A\cong A\times \{a\}$. Thus $A$ can be embedded in $\mathbb{R}$. Therefore $A$ is metrizable.

Thank you

Ivo Terek
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Amr
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3 Answers3

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No, no such space exists. Suppose $A$ is a topological space such that $A\times A\cong \mathbb R$, with the Euclidean induced topology on $\mathbb R$. If $A$ is disconnected then so is $A\times A$, but that would contradict the connectivity of $\mathbb R$, so it follows that $A$ is connected. But, since $A$ is connected it follows that $A\times A$ with a single point removed is still connected. However, $\mathbb R$ with a single point removed is not connected. Contradiction.

Ittay Weiss
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    A very easy yet fascinating argument! +1 upvote, of course. – Sangchul Lee Mar 12 '13 at 09:48
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    $A$ is the continuous image of $\mathbb{R}$ under projection so must be connected. This simplifies the argument a little. – Henno Brandsma Mar 12 '13 at 14:07
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    Hi Ittay. Maybe this question is just triviality, but could you clarify why $A\times A$ with a single point removed is still connected? I know if $A=\mathbb{R}$ then one can either produce arguments with topological dimensions or use path connectedness. But for any connected space $A$ I can't see how it follows immediately. – T. Eskin Mar 12 '13 at 18:25
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    @ThomasE. you can express it as a union of connected parts with connected intersections. Alternatively, the whole argument can be repeated with path connectedness instead of connectedness. Henno's argument also shows that $A$ must be path connected, since it's a continuous image of $\mathbb R$. – Ittay Weiss Mar 12 '13 at 18:37
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If $A \times A \simeq \mathbb{R}$, then $A$ has to be connected and it is homeomorphic to a subspace of $\mathbb{R}$, so $A$ is homeomorphic to an interval. Therefore, either $A \times A \simeq [0,1]^2$ or $A \times A \simeq (0,1)^2 \simeq \mathbb{R}^2$ or $A \times A \simeq [0,1)^2$. Thus, $A \times A$ can't be homeomorphic to $\mathbb{R}$.

Seirios
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    You should probably say some things about why those spaces are not homeomorphic to $\mathbb{R}$. The easiest argument is probably that none of them can be disconnected by removing a point, which is just Ittay Weiss' argument. – Qiaochu Yuan Mar 13 '13 at 02:09
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This was intended as a comment on the answers above, but I'm a new user, so I don't have that as an option, so I'll need to post it as a new answer instead.

The other answers here show that $\mathbb{R}$ cannot be written as a "square" of sorts. The situation is not unlike that of the real number -1, which cannot be written as a square of real numbers.

But one can extend the real numbers to create a square root of $-1$. Similarly, one could extend the notion of topological space to create a "square root" of $\mathbb{R}$. One way to do so, in analogy with the complex numbers, would be as follows. Define a "complex topological space" to be an ordered pair $(A,B)$ of topological spaces. Define the direct product of two such ordered pairs by

$(A,B) \times (C,D)$ equals (disjoint union of $A\times C$ and $\mathbb{R}\times B \times D$, disjoint union of $A\times D$ and $B\times C$)

Identify a topological space $Y$ with the ordered pair $(Y, \varnothing)$.

Then $(\varnothing$, one-point set$)\times (\varnothing$, one-point set) = ($\mathbb{R}$, empty set).

mez
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Mike Krebs
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