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I have a hard time believing that there can exist a bijection $f:\Bbb R^2\to \Bbb R$.

I just cannot get around my intuition that a one-to-one map of a one-dimensional space (or interval) must also be one-dimensional.

I am not a mathematician. I am interested in the topic since it has potential relevance (albeit not necessarily importance) for economic theory I am working on.

I have looked at the construction of a bijection here: https://math.stackexchange.com/a/183383. And I have also noted the comment made here that such a bijection cannot be continuous: https://math.stackexchange.com/a/43098/429504

But, does not the bijection offered in the first thread preserve convergent sequences?

Let's look at a bijection between the unit square and unit interval as in the first thread. Consider a sequence of real numbers $x_1,x_2,...$ that converges to a real number $x \in [0,1]$. Then there exists, for any natural number $m$ a natural number $k$ so that every real number $x_n, n > k$, in the sequence agrees with $x$ on the first $m$ digits after the decimal point. Since $m$ is arbitrary, the proposed bijection will create a sequence $f(x_n) \in [0,1]^2$ so that both elements of that two-vector never changes any of its $l$ first digits for $n > k$. It follows from the definition of the bijection that $l$ can be made arbitrarily large by making $m$ sufficiently large which in turn is achieved by making $k$ sufficiently large.

Hence, the bijection $f$ seems to always map a convergent sequence in $\Bbb R$ to a convergent sequence in $\Bbb R^2$. Does not this mean that the proposed bijection is continuous, leading to a contradiction with the comment in the second thread?

(This is my first post here and I am not a mathematician, so my apologies for what may be a very clumsy post.)

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Contrarian
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  • But you’re assuming that the map $,f$ is continuous. The only possible such maps, in fact, will be strongly discontinuous. – Lubin Mar 27 '17 at 01:37
  • I am referring to the map $f$ proposed here http://math.stackexchange.com/a/183383 that I have also seen proposed elsewhere, but yes, that map seems to be continuous, or am I missing something? – Contrarian Mar 27 '17 at 01:45
  • That map is most certainly not continuous. – Lubin Mar 27 '17 at 01:47
  • $\ldots$ and (somewhat less to the point) it also doesn't map $\mathbb{R}^2$ onto $\mathbb{R}$ – Martin Rattigan Mar 27 '17 at 02:49

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Digits and convergence do not behave well together. If $x_1,x_2,\ldots$ converges to $x$, there may never be an $n$ so that even the first digit of $x_n$ is the same as that of $x$. Consider, for example, the sequence $0.9$, $0.99$, $0.999$, $\ldots$, and its partner $1.1$, $1.01$, $1.001$, $\ldots$. These two sequences converge to the same point, but the construction outlined in what you linked sends them to wildly different places. In fact, a situation like this can be arranged around every rational point - which means that this function has densely many discontinuities.

Reese Johnston
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It is not a mapping of the spaces but a mapping of the sets. A 1-1 function from R^2 to R would be simply to go back and forth choosing digits from each number. Ie: f(.111111...,.22222...)=.12121212121212...

Jacob Wakem
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  • Thank you! It was my thinking of this as a mapping of spaces that lead my intuition astray. I was imagining the result has more implications than it actually does. Thinking of it as a mapping of sets, it seems rather trivial and believable. The result is just not implying as much as I imagined it did. – Contrarian Mar 28 '17 at 16:42
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There is a bijection between $\mathbb{R}$ and $(0,1)$ and a bijection between $(0,1)$ and a Cantor set $K$.
The first map is given by $x\mapsto \frac{1}{2}+\frac{1}{\pi}\arctan(x)$ and the second map takes the canonical (i.e. without an infinite tail of digits $1$) base-$2$ representation of $x\in(0,1)$ and does the following: $$ 0.010011001_2 \mapsto 0.020022002_3 $$ (i.e. replace every digit $1$ with a $2$ and consider the resulting string as a base-$3$ representation).
There is a bijection between $K\times K $ and $K$: $$ 0.\color{red}{002202},\quad 0.\color{blue}{202002}\quad \Longleftrightarrow \quad 0.\color{red}{0}\color{blue}{2}\color{red}{0}\color{blue}{0}\color{red}{2}\color{blue}{2}\color{red}{2}\color{blue}{0}\color{red}{0}\color{blue}{0}\color{red}{2}\color{blue}{2}$$ given by zipping or un-zipping ternary representations. There is a bijection between $K\times K$ and $(0,1)\times (0,1)$ and a bijection between $(0,1)\times(0,1)$ and $\mathbb{R}\times\mathbb{R}$, hence there is a bijection betwen $\mathbb{R}$ and $\mathbb{R}\times\mathbb{R}$.

We took a difficult path, since such bijection simply exists due to the Cantor-Schroeder theorem: an injective map from $\mathbb{R}$ to $\mathbb{R}\times\mathbb{R}$ is trivial and an injective map from $\mathbb{R}\times\mathbb{R}$ to $\mathbb{R}$ is given, for instance, by a "zipping procedure" similar to the one outlined above.

But of course there is no continuous bijection since $\mathbb{R}$ and $\mathbb{R}^2$ have different topological behaviours: if we remove a point from $\mathbb{R}$ we disconnect it, while the same does not happen to $\mathbb{R}^2$. Since connection is a topological property (it is preserved by continuous maps), there cannot be any continuous bijection between $\mathbb{R}$ and $\mathbb{R}^2$.

It is worth mentioning that the Peano curve is a continuous and surjective function from $[0,1]$ to $[0,1]^2$, hence there is a continuous and surjective function $f:\mathbb{R}\mapsto\mathbb{R}^2$, but it is not injective.

Jack D'Aurizio
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  • Thank you for your detailed answer, I have to spend some time to grok it as I need to read up on this stuff. Thank you for mentioning the Peano curve which is news to me ;) It is intuitive to me that when you divide the unit square into a grid with all component squares of non-zero length, you can draw an injective curve over the unit interval that hits all the squares. It's the limit of that being injective I am uncomfortable with. – Contrarian Mar 27 '17 at 15:14
  • @Contrarian: as a matter of fact the limit of such curves is not injective, but it is surjective. – Jack D'Aurizio Mar 27 '17 at 15:16
  • Yes indeed, and it is the injective part that I am particularly interested in and have a hard time visualizing that is possible. Thanks again! – Contrarian Mar 27 '17 at 15:25
  • @Contrarian: a continuous and injective map would give that $\mathbb{R}$ and $\mathbb{R}^2$ have the same topological structure, that is disproved by the paragraph about connection. Injectivity (from $\mathbb{R}$ to $\mathbb{R}^2$) + continuity is not possible. Surjectivity + continuity is possible. – Jack D'Aurizio Mar 27 '17 at 15:48
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This might help clear things up: Let $B$ denote the set of binary sequences, i.e., those sequences $(a_n)$ for which $a_n \in \{0,1\}$ for each $n.$ I'm hoping you are familiar with the fact that $B$ and $\mathbb R$ have the same cardinality. In this cleaner setting, where we don't have to worry about decimal expanisons ending in all $9$'s, the map from $B\times B \to B,$ given by $((a_n),(b_n)) \to (a_1,b_1,a_2,b_2, \dots),$ is a bijection. Therefore $\mathbb R\times \mathbb R$ and $\mathbb R$ have the same cardinality.

zhw.
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