Trying to define $\mathbb{R}^{0.5}$ topologically 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
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.
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).