Which (possibly implicit) assumptions and conclusions (that later turned out false) made it hard for Cantor to believe that there is a bijection between the unit interval $[0,1] \subset \mathbb{R}$ and the unit square $[0,1]^2$, i.e. $|[0,1]| = |[0,1]^2|$?

It was all clear to him that there is a bijection between $\mathbb{Z}$ and $\mathbb{Q} \equiv \mathbb{Z}^2$ and therefore between $\mathbb{Q}$ and $\mathbb{Q}^2$, i.e. $|\mathbb{X}| = |\mathbb{X}^2|$ for some countable infinite sets (even when not enumerable in natural order). He also knew that $|\mathbb{Q}^\sqrt{}| = |\mathbb{Q}^{\sqrt{}} \times \mathbb{Q}^{\sqrt{}} |$ for the "Euclidean" numbers (mainly because $\mathbb{Q}^\sqrt{}$ is countable) and probably that $|\mathbb{X}| = |\mathbb{X}^2|$ for all countable sets $\mathbb{X}$.

But his assumption must have been that the latter doesn't necessarily hold for uncountable sets. If he had not thought so, he would not have been surprised to find that there is a bijection between $[0,1]$ and $[0,1]^2$, but he seemed to be so:

"I see it, but I don't believe it."

But for what specific reasons did he believe that $|\mathbb{X}| = |\mathbb{X}^2|$ does not necessarily hold for uncountable sets $\mathbb{X}$?

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    $\begingroup$ I recall reading that he had difficulty accepting this because he -- like most others in his time -- was used to dealing with continuous mappings, and of course there is no continuous bijection between the segment and the square: that would kill a whole lot of the field of topology. But I don't have a source for this I can find back. $\endgroup$ – Mees de Vries Sep 12 '18 at 8:44
  • $\begingroup$ But already Cantor's mapping from $\mathbb{Z}$ to $\mathbb{Q}$ was somehow "dicontinous" (not mapping nearby arguments to nearby values), and he was used of (or even invented?) such mappings. And finally his proof of $|[0,1]| = |[0,1]^2|$ was analogous to his proof of $|\mathbb{Z}| = |\mathbb{Q}|$ (by diagonalization). $\endgroup$ – Hans-Peter Stricker Sep 12 '18 at 8:59
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    $\begingroup$ @MeesdeVries - correct. See Dedekind's comment to Cantor's proof : "If a reciprocally unique and complete correspondence between the points of a continuous domain $A$ of $a$ dimensions on the one hand and the points of a continuous domain $В$ of $b$ dimensions on the other is possible, then this correspondence is necessarily completely discontinuous if $a$ and $b$ are unequal." Dauben, page 57. $\endgroup$ – Mauro ALLEGRANZA Sep 12 '18 at 9:21
  • $\begingroup$ But this doesn't imply that there isn't such a correspondence (which Cantor seemed to believe first). And as I said: Cantor was used to (somehow) discontinous mappings. (And: Dedekind's statement remained true.) $\endgroup$ – Hans-Peter Stricker Sep 12 '18 at 9:35
  • $\begingroup$ What is X and $Q^\sqrt?$ $\endgroup$ – William Elliot Sep 12 '18 at 9:38

It seems to me that the answer to your question is already contained in the article you linked to. Cantor says that he originally didn't expect this because it was widely believed that it takes $n$ coordinates to specify a point in an $n$-dimensional manifold. (He also says that others held this to be self-evident whereas he believed it required a proof.) By the way, the article argues that his remark "I see it, but I don't believe it." doesn't actually, as you seem to imply, refer to the result but to the proof.

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  • $\begingroup$ But the same arguments goes for $\mathbb{Z}$ and $\mathbb{Q}$: you need 2 integers to specify a rational number. $\endgroup$ – Hans-Peter Stricker Sep 12 '18 at 10:06
  • $\begingroup$ @HansStricker: That's not the same argument. You need two integers to specify an element of $\mathbb Z_3\times\mathbb Z_5$, but it's pretty obvious that you can use one integer instead. So there are different levels of belief that two numbers can't be expressed in one. The belief was stronger for $\mathbb R^2$ than for $\mathbb Z^2$ because of the continous structure of $\mathbb R^2$. Search for "manifold" and "continuous" in the article you linked to to find indications of this in both Cantor's and Dedekind's letters. $\endgroup$ – joriki Sep 12 '18 at 10:39
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    $\begingroup$ @HansStricker: While this is not supported by the article, I would suspect that naive notions of "area" and "volume" were also partly to blame. Unless you're steeped in the cardinality paradigm that Cantor ushered in, it seems easier to think that some set of dimensionless points ($\mathbb Z^2$) could be "as much" as some other set of dimensionless points ($\mathbb Z$) than that something having volume ($\mathbb R^3$) or area ($\mathbb R^2$) could be "as much" as something having only length ($\mathbb R$). $\endgroup$ – joriki Sep 12 '18 at 10:41
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    $\begingroup$ Thanks a lot for your first comment, concerning levels of belief. $\endgroup$ – Hans-Peter Stricker Sep 12 '18 at 18:04

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