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Is there any such isomorphism? Or between any closed interval on $\mathbb{R}$ and any closed simply connected subset of $\mathbb{R}^2$? If so, how is it expressed and does it preserve any structure?

(Disclaimer: I'm not a mathematician by discipline so hopefully my question is actually meaningful!)

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Isomorphism of what? Sets? Topological spaces? –  Qiaochu Yuan Oct 12 '11 at 14:56
Isomorphism means literally similarity of structure. So in order to say that such isomorphism exists or not, first we need to know what sort of structure needs to be preserved. –  Asaf Karagila Oct 12 '11 at 15:04
As for your second question, $[0,1]$ sits inside $[0,1]^2$ as $[0,1]\times\{0\}$, which is a closed connected subset. –  Grumpy Parsnip Oct 12 '11 at 15:09
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4 Answers

up vote 4 down vote accepted

An "isomorphism" between to mathematical "structures" $A$ and $B$ is a bijective map $\phi$ between the two underlying sets to begin with. If these structures are of an algebraic nature one would, e.g., require that $\phi(x)+\phi(y)=\phi(z)$ whenever $x+y=z$. If $A$ and $B$ are metric spaces with distance functions $d_A$ and $d_B$ then one would require that $d_B\bigl(\phi(x),\phi(y)\bigr)=d_A(x,y)$ for all $x$, $y\in A$, etc.

Now you are asking about $A=[0,1]$ and $B=[0,1]^2$. It is indeed possible to define a bijective map between these two sets, as sketched in other answers. But these sets have additional structure, e.g., they are topological spaces. Intuitively it is obvious that as topological spaces they are not isomorphic, but an actual proof is difficult, since it would be an "impossibility proof" (note that it took mankind 2000 years to prove that you cannot trisect an angle with ruler and compass). One has to find a certain topological property that $A$ has (e.g., it is "one-dimensional", whatever that means), which $B$ does not have.

O.k., here is a proof, but it uses various concepts whose validity would have to be established beforehand: Consider the point $M:={1\over2}\in A$. Any continuous path in $A$ connecting $P:={1\over4}$ with $Q:={3\over4}$ must pass through $M$ – this is the intermediate theorem of calculus. But given any point $M\in B$ and any two points $P$, $Q\in B$ you can find a continuous path in $B$ connecting $P$ and $Q$ without passing through $M$. This proves that $A$ and $B$ are not topologically equivalent.

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As Qiauchu Yuan said above, it depends on the level of structure you want your map to preserve. If you just want to ask "is there a map that takes points to points and is one-to-one and onto?", then the answer is yes. This is the same thing as saying they have the same cardinality as sets, and one such map would take the point $0.d_1 d_2 d_3 d_4 \ldots$ to the point $(0. d_1 d_3 \ldots , 0.d_2 d_4 \ldots)$. If you want to ask, "is there a map that takes open sets to open sets (under the standard interval topologies) and is one-to-one and onto?", the answer is no.

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This map is well-known to be mildly problematic; there are issues with the non-uniqueness of decimal representations owing to the possibility of repeating $9$s. –  Qiaochu Yuan Oct 12 '11 at 15:12
Cantor-Bernstein easily (but nonconstructively) fixes that, however. –  Henning Makholm Oct 12 '11 at 15:24
@Henning: What notion of "constructive" are you using? All the proofs of Cantor-Bernstein I know directly spit out a definition of the desired bijection. –  user83827 Oct 12 '11 at 15:32
@ccc, you're right -- "nonconstructive" was too strong a word to use there. I was trying to express that Cantor-Bernstein does not (in general) lead to an explicit form from which the partner of one element can be calculated concretely. –  Henning Makholm Oct 12 '11 at 15:41
@Arthur: A space-filling curve from an interval to a square is not injective. If it was, the square would be homeomorphic to the interval. –  LostInMath Oct 12 '11 at 15:55
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If you are talking about topology, then, when you "remove" the point $\frac{1}{2}$ from $[0,1]$, then it is no longer connected. But when you "remove" the "corresponding" point from $[0,1]^2$, it remains connected.

That is, letting $a = f(1/2)$, where $f$ is your "isomorphism", then, the restriction of $f$ to $[0,1] \setminus \left\{\frac{1}{2}\right\}$ should be an "isomorphism" to its image $[0,1]^2 \setminus \left\{f\left(\frac{1}{2}\right)\right\}$. But this cannot happen, since one is disconnected while the other is not.

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I like your answer :-) (+1). –  robjohn Oct 12 '11 at 18:42
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I assume that you are looking at $[0,1]$ and $[0,1]^2=[0,1]\times[0,1]$ as subsets of $\mathbb{R}$ and $\mathbb{R}^2=\mathbb{R}\times\mathbb{R}$ with their standard topologies as metric spaces. I also assume that by "isomorphism", you mean a bi-continuous function between them.

My answer to a question about cardinalities gives a pair of injections between $[0,1)$ and $[0,1)\times[0,1)$ and then cites the Cantor-Bernstein-Schroeder Theorem to get a bijection between $[0,1)$ and $[0,1)\times[0,1)$. Breaking the "finite when possible" rule there and writing $1$ as $$ \sum_{k=1}^\infty 2^{-k} $$ we get a bijection between $[0,1]$ and $[0,1]\times[0,1]$. However, this map cannot be bi-continuous, so it is not an isomorphism.

Suppose we had a bicontinuous map $f:[0,1]\leftrightarrow[0,1]\times[0,1]$. Since $f^{-1}$ is continuous, it maps connected sets to connected sets. Let $f\left(\frac{1}{2}\right)=x\in[0,1]\times[0,1]$. Since $f$ is a bijection, $f^{-1}([0,1]\times[0,1]\setminus \{x\})=[0,1]\setminus\{\frac{1}{2}\}$. However, $[0,1]\times[0,1]\setminus \{x\}$ is connected, but $[0,1]\setminus\{\frac{1}{2}\}$ is not.

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