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There are a lot of websites and forums, which explain that there is a bijection between $\mathbb{R}$ and $\mathbb{R}^2$, and even give some bijections. (By the way: Can you generalize it? since it works with the natural Numbers and the real numbers, does there exist a bijection between any infinite set $X$ and $X^2$) On some websites there is claimed that there doesn't exist a continuous bijection. But how would I approach such a proof? (I don't need a complete proof,... just a starting point because I have absolutely no idea how to begin this.)

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  • $\begingroup$ Yes, for any infinite set $X$ there is a bijection from $X$ to $X^2$. $\endgroup$ – Arthur Nov 5 '15 at 12:31
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    $\begingroup$ Consider Netto's theorem; If $f$ represents a bijective map from an $m$-dimensional smooth manifold $\mu_m$ onto an $n$-dimensional smooth manifold $\mu_n$ and $m \neq n$, then $f$ is necessarily discontinuous. $\endgroup$ – Lionel Ricci Nov 5 '15 at 12:33
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    $\begingroup$ @Arthur Only with the axiom of choice, though. In fact, the statement that $|X|=|X|^2$ is true for all $X$, is equivalent to the axiom of choice. See here. $\endgroup$ – Akiva Weinberger Nov 5 '15 at 12:34
  • $\begingroup$ @LionelRicci : Since your comment essentially answered the question, you might consider un-deleting the answer (or add more information to it if you think it is a bit short). $\endgroup$ – user99914 Nov 5 '15 at 12:40
  • $\begingroup$ Related question here $\endgroup$ – user99914 Nov 5 '15 at 13:58
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Easy direction: there is no continuous bijection from $\mathbb R^2$ onto $\mathbb R$. For suppose $f : \mathbb R^2 \to \mathbb R$ is continuous. Choose one point $a \in \mathbb R^2$. The deleted set $\mathbb R^2 \setminus \{a\}$ is connected. A continuous image of a connected set is connected. So $f\big(\mathbb R^2 \setminus \{a\}\big)$ is a connected subset of $\mathbb R$. In particular, it is not the deleted set $\mathbb R \setminus \{f(a)\}$, since that is disconnected. So either $f$ is not injective or not surjective.

Note The same argument shows there is no continuous bijection from an open ball in $\mathbb R^2$ onto a subset of $\mathbb R$. You just have to make sure to choose $a$ that does not map to the maximum or minimum of the image.

Hard direction: there is no continuous bijection from $\mathbb R$ onto $\mathbb R^2$. This will use the Baire Category Theorem. Let $f : \mathbb R \to \mathbb R^2$ be continuous and injective. $\mathbb R$ is the countable union of compact sets $[-n,n]$. I will show that $f\big([-n,n]\big)$ has empty interior. Then $f(\mathbb R)$ is a countable union of closed sets with empty interior, so by the Baire Category Theorem, it is not $\mathbb R^2$.

So, why does $K_n:=f\big([-n,n]\big)$ have empty interior? Suppose it has nonempty interior. Since $[-n,n]$ is compact and $\mathbb R^2$ is Hausdorff, the continuous bijection $f$ from $[-n,n]$ onto $K_n$ is a homeomorphism. So the restriction of $f^{-1}$ to an open ball contained in $K_n$ would be a homeomorphism from an open ball in $\mathbb R^2$ onto a subset of $\mathbb R$. Contradiction.

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For the question about generalization, consider Netto's theorem;

If $f$ represents a bijective map from an $m$-dimensional smooth manifold $\mu_m$ onto an $n$-dimensional smooth manifold $\mu_n$ and $m \neq n$, then $f$ is necessarily discontinuous.

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