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Consider the following statement:

Prove that it is possible to write $\Bbb R$ as a union $\Bbb R= \bigcup_{i\in I} A_{i}$ where $A_{i} \cap A_{j}= \emptyset$ if $i\neq j$, $i,j \in I$,and such that each $A_{i}$ and $I$ are uncountable sets.

Thanks for Kyle Gannon who gives a constructive proof (The real numbers as the uncountably infinite union of disjoint uncountably infinite sets):

Since $|\mathbb{R}| = |\mathbb{R} \times \mathbb{R}| $, there exists a bijection $f$ from $\mathbb{R} \to \mathbb{R} \times \mathbb{R} $. Then $\mathbb{R} = \bigcup_{a \in \mathbb{R}} f^{-1}(\mathbb{R},a)$ where $f^{-1}(\mathbb{R},a) = \{b \in \mathbb{R}: f(b) = (c,a)$ for some $c \in \mathbb{R} \}$.

People say that each $f^{-1}(\mathbb{R},a)$ is the preimage of one vertical line in the plane. But it seems to me that each $f^{-1}(\mathbb{R},a)$ should be the preimage of one horizontal line in the plane. For example, $f^{-1}(\mathbb{R},\sqrt2)$ is the set $\lbrace b \in \mathbb{R}: f(b) = (c,\sqrt 2)$ for some $c \in \mathbb{R} \}$. Is my understanding correct?

To be precise, the set of all points in one single horizontal line is a subset of the codomain $ \Bbb R^2$. My question is whether the preimage of this set (the set of all points in that horizontal line) with respect to the map $f$ is one such $f^{-1} (\Bbb R, a)$.

Peter Taylor
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Y.X.
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  • Well that would be true if $f:\mathbb R \to \mathbb R$ which it isn't. It would be better to treat this without vertical and horizontal lines, as they aren't defined at all nor are they needed. All you need to know is that for $f^{-1}$ the preimage is $\mathbb R \times \mathbb R$. – Sentinel135 Mar 13 '17 at 02:27
  • There is no function $f^{-1}$ involved at all. The notation $f^{-1}$ is just for preimage. – Y.X. Mar 13 '17 at 02:39
  • "Since $|\mathbb R|=|\mathbb R\times \mathbb R|$ there exists a bijection $f$ from $\mathbb R\to \mathbb R\times \mathbb R$." You defined the function $f$ here, and the preimage of $f$ is $\mathbb R$ and the image is $\mathbb R\times \mathbb R$ since $f$ is a bijection. So for $f^{-1}$, which exists because $f$ is bijective, the preimage of $f^{-1}$ is the same as the image of $f$ hence $\mathbb R\times \mathbb R$ . – Sentinel135 Mar 13 '17 at 02:47
  • The set of all points in one single horizontal line is a subset of the codomain $ \Bbb R^2$. My question is whether the preimage of this set (the set of all points in that horizontal line) with respect to the map $f$ is one such $f^{-1} (\Bbb R, a)$. – Y.X. Mar 13 '17 at 02:55
  • If you think of $f^{-1} (\Bbb R, a)$ as the inverse function with input $(\Bbb R, a)$, that is fine because the input just means $a$ (the y-coordinate) is fixed and the x-coordinate is unfixed. Isn't that one single horizontal line? – Y.X. Mar 13 '17 at 03:02
  • Ok so you are asking for the preimage of $f$ for a subset of $f^{-1}(\mathbb R,a)$? To answer that question we need to know more about the function. as in what maps to what and how. keep in mind it would most likely be a feather dusting on $\mathbb R$ or a uncountably infinite set of disjointed points on $\mathbb R$. – Sentinel135 Mar 13 '17 at 03:02
  • No I am not asking for the preimage of $f$ for a subset of $f^{-1} (\Bbb R, a)$. $(\Bbb R, a)$ is a subset of $\Bbb R^2$. $f^{-1} (\Bbb R, a)$ is the preimage of $(\Bbb R, a)$ with respect to $f$. That is why I am saying there is no inverse function involved at all. – Y.X. Mar 13 '17 at 03:06
  • No, because 1) $(\mathbb R, a)$ is a set of points and not a function. and 2) the preimage of $f$ is $\mathbb R$ and not in $\mathbb R^2$. – Sentinel135 Mar 13 '17 at 03:13
  • I never said $(\Bbb R, a)$ is a function. I said in my previous comment that it is a subset of $\Bbb R^2$. It does not make sense to talk about "the preimage of $f$". By definition of the preimage, it makes sense to talk about the preimage of a subset of the codomain with respect of $f$, which is a subset of the domain. Check this:https://en.wikipedia.org/wiki/Image_(mathematics) – Y.X. Mar 13 '17 at 03:18
  • Yes but any preimage of $f$ lives in $\mathbb R$ and not in $\mathbb R^2$. So looking at $(\mathbb R^2,a)$ as a preimage of $f$ makes no sense. However this doesn't seem to be your issue. Your issue is if the set $(\mathbb R,a)$ is a vertical line or a horizontal one. This is a horizontal line with the equation $y=a$. – Sentinel135 Mar 13 '17 at 03:57

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You are correct: the set $f^{-1}(\mathbb{R},a)$ is the preimage of a horizontal line in the plane, not a vertical line in the plane.

(I'm not sure who the "people" are who claim otherwise--no one said anything of the sort in the post you linked to.)

Eric Wofsey
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