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Let's say $f(x,y)$ is a continuous function. $x$ and $y$ can be any real numbers. Can this function have one unique value for any two different pairs of variables? In other words can $f(a,b) \neq f(c,d)$ for any $a$, $b$, $c$, and $d$ such that $a \neq c$ or $b \neq d$? I don't think there can at least not if the range of $f$ is within the real numbers. Could someone please offer a more formal proof of this or at least start me off in the right direction.

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Assume $f : \mathbb{R}^2 \to \mathbb{R}$ is continuous and injective. Then for each fixed $y$, the function $x \mapsto f(x,y)$ is monotonic. Its image is some interval, and in particular contains a rational number. None of these points can be re-used for some other $y$. So $y$ can't be drawn from an uncountable set, since the rationals are countable. But $\mathbb{R}$ is uncountable...

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    $\begingroup$ @goblin Can you give a counterexample? Continuous, injective functions from $\mathbb{R}$ to itself are monotone, so the only possibility is that $x \mapsto f(x,y)$ is not continuous. $\endgroup$ – Ian Jan 19 '16 at 1:59
  • $\begingroup$ @HenningMakholm Thanks for the clarifying edit. $\endgroup$ – Ian Jan 19 '16 at 2:51
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    $\begingroup$ But if the range is the complex numbers it can work right? $\endgroup$ – tox123 Jan 19 '16 at 3:24
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    $\begingroup$ @tox123: Yes, of course: $(x,y)\mapsto x+iy$. $\endgroup$ – hmakholm left over Monica Jan 19 '16 at 8:28
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If $f$ takes values in $\mathbb R$, then no, $f$ cannot be injective.

To see this, consider the function $$ g(t) = f(\cos t, \sin t) $$ which is continuous because it is a composition of continuous functions. If $f(1,0)=f(-1,0)$ then we know $f$ is not injective. Otherwise, since $g(0)=g(2\pi)$, by the intermediate value theorem the value $\frac{g(0)+g(\pi)}{2}$ must be attained for some $t_0\in(0,\pi)$ as well as for some $t_1\in(\pi,2\pi)$. But $(\cos t_0,\sin t_0)$ is a different point than $(\cos t_1, \sin t_1)$. so in this case $f$ is not injective either.


On the other hand, there are continuous injections $\mathbb Q^2\to\mathbb Q$. Namely, $(x,y)\mapsto x+\sqrt2 y$ injects $\mathbb Q^2$ continuously into $\mathbb Q+\sqrt2 \mathbb Q$, and this set is a countable dense linear order without first and last element and therefore order-isomorphic -- and thus homeomorphic -- to $\mathbb Q$.

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  • $\begingroup$ That is not a proof. $\endgroup$ – AJY Jan 19 '16 at 1:59
  • $\begingroup$ @AJY: Which step do you find unconvincing? $\endgroup$ – hmakholm left over Monica Jan 19 '16 at 2:00
  • $\begingroup$ OP asked to show that a continuous bijection did not exist, and you provided two examples of functions which did not do it. $\endgroup$ – AJY Jan 19 '16 at 2:01
  • $\begingroup$ @AJY: Um, no. I show that an arbitrary continuous function $f:\mathbb R^2\to\mathbb R$ is non-injective. $\endgroup$ – hmakholm left over Monica Jan 19 '16 at 2:03
  • $\begingroup$ My apologies. I misread. $\endgroup$ – AJY Jan 19 '16 at 2:03
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No, because the continuous image of a connected set is connected. If $f: \mathbb{R}^{2} \to \mathbb{R}$ is a continuous bijection, then we can consider $f( \mathbb{R}^{2} \setminus \{ 0, 0 \} ) = \mathbb{R} \setminus \{ f(0, 0) \}$. Let $c = f(0, 0)$. Since $\mathbb{R}^{2} \setminus \{ (0, 0) \}$ is connected, so must be $\mathbb{R} \setminus \{ f(0, 0) \} = ( - \infty, c) \cup (c, + \infty)$, which it obviously isn't. Thus either $f$ is not continuous, or it is not a bijection.

EDIT: Commenter pointed out that the OP only supposed $f$ was injective. A similar method holds. Again, $f (\mathbb{R}^{2})$ is a connected subset of $\mathbb{R}$, and so is an interval, since all connected subsets of $\mathbb{R}$ are intervals (though its endpoints may be infinite). So suppose that the interval $(a, b) \subseteq f(\mathbb{R}^{2})$. Let $a < c < b$, and let $\mathbf{x} = (x_{1}, x_{2})$ be such that $f(\mathbf{x}) = c$. Then again, $\mathbb{R}^{2} \setminus \{ \mathbf{x} \}$ is connected, so $f( \mathbb{R}^{2} \setminus \{ \mathbf{x} \} )$ is connected. But we can consider that $$f( \mathbb{R}^{2} \setminus \{ \mathbf{x} \} ) = ( (- \infty, c ) \cap f(\mathbb{R}^{2} \setminus \{ \mathbf{x} \})) \cup ( (c, + \infty) \cap f(\mathbb{R}^{2} \setminus \{ \mathbf{x} \})) ,$$ and so is not connected. Thus either $f$ is not injective, or its not continuous.

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  • $\begingroup$ Strictly speaking the OP asked only for a continuous injection, not necessarily a bijection. $\endgroup$ – hmakholm left over Monica Jan 19 '16 at 2:09
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You are asking for a continuous injective function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$. Consider $C=[-1,1] \times [1,1]$. $f|_C$ is still injective and continuous. Since it is defined on a compact set, it is a homeomorphism with its image (note that $\mathbb{R}$ is Hausdorff). But $C$ is also connected, hence the image must be a closed interval $[a,b]$. This would tell us that there exists a homeomorphism between $[-1,1]$ and $[-1,1] \times [1,1]$, which is false (for example, you can take off one point of $[-1,1]$ and this will be not connected, whereas taking off one point of $C$ leaves a connected set).

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Take any pair of distinct points $p,q$ of $\Bbb R^2$ and link them by two disjoint (except in the end points) arcs. For instance take $p=(1,0)$ and $q=(-1,0)$, linked by the upper and lower halves of the unit circle. Now by hypothesis $f(p)\neq f(q)$, and the restriction of $f$ to either arc is a continuous function (of the parameter for the arc). By the intermediate value theorem, every value in the interval between $f(p)$ and $f(q)$ is taken by $f$ on either of the arcs, but this contradicts the assumed injectivity.

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  • $\begingroup$ I see this is basically the same as the answer by Henning Makholm, but I just wanted to make the "two arcs" argument stand out more clearly. $\endgroup$ – Marc van Leeuwen Jan 19 '16 at 5:20

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