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Consider the surjection $[0,2\pi)*\left(\mathbb{R}^2\cup\ell_\infty\right)\rightarrow L$ (such that $L$ is the set of all lines determined by the binary operation $*$ between a point and an angle). Therefore, $\left|L\right|\leq\left|[0,2\pi)*\left(\mathbb{R}^2\cup\ell_\infty\right)\right|=\mathfrak{c}*(\mathfrak{c}^2+\mathfrak{c})=\mathfrak{c}*\mathfrak{c}$. Is $\mathfrak{c}*\mathfrak{c}=\mathfrak{c}$? How would I show this?

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Asaf Karagila
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UNOwen
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2 Answers2

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Yes it is. There is a general proof of the statement $|A\times A|=|A|$ for every infinite set $A$ that relies on Axiom of Choice. But for this particular case you can do a much easier constructive proof of $|\mathbb{R}\times\mathbb{R}|=|\mathbb{R}|$:

  • Obviously $|\mathbb{R}|\leq |\mathbb{R}\times\mathbb{R}|$.
  • Let's see that there is a surjective map from $[0,1]$ to $[0,1]\times[0,1]$, and thus $|\mathbb{R}\times\mathbb{R}|=|[0,1]\times[0,1]|\leq |[0,1]|=|\mathbb{R}|$. Let $x:=0.x_1x_2x_3\ldots$ denote the decimal expansion of $x$. In case of $x$ having more than one, consider the shorter one. Consider the map: $h:[0,1]\longrightarrow[0,1]\times[0,1]$ $$x=0.x_1x_2x_3\ldots\longmapsto h(x):=(0.x_1x_3x_5\ldots,0.x_2x_4x_6\ldots)$$ This map is surjective. If we take $(0.x_1x_2x_3\ldots,0.y_1y_2y_3\ldots)\in[0,1]\times[0,1]$, we have that: $$h(0.x_1y_1x_2y_2x_3y_3\ldots)=(0.x_1x_2x_3\ldots,0.y_1y_2y_3\ldots)$$ Proving the claim.
R.V.N.
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  • Thank you for your answer. Does $\times$ denote a binary operation? If so, I understand fully. Also, what would $2^\mathfrak{c}-\mathfrak{c}$ be (i.e. the cardinality of the power set of all points minus the cardinality of all lines)? – UNOwen Feb 24 '21 at 10:34
  • I posted an answer to your question as a comment because of it's length. About the $\times$ symbol, it denotes cartesian product ($\mathbb{R}\times\mathbb{R}=\mathbb{R}^2$). – R.V.N. Feb 24 '21 at 16:07
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I will answer here to your comment question. Again, there are general results that allow us to directly proof that $2^\mathfrak{c}-\mathfrak{c}$ is $2^\mathfrak{c}$, but their proof is not easy at all. So I've been thinking for a long time for a constructive proof of this particular case and I came to this, but there might be some failure on my proof. In that case let me know.

To prove this it suffices to prove that there is a bijection between the sets $2^\mathbb{R}\setminus\{\{x\}:x\in\mathbb{R}\}$ and $2^\mathbb{R}$. Proving that $|2^\mathbb{R}\setminus\{\{x\}:x\in\mathbb{R}\}|\leq|2^\mathbb{R}|$ is direct, so let's focus just on the other one.

Consider the following map $f:2^\mathbb{R}\longrightarrow2^\mathbb{R}\setminus\{\{x\}:x\in\mathbb{R}\}$: $$S\longmapsto f(S):=\begin{cases}S &\text{ if } S \text{ is infinite} \\ S\cup \{2·s_k\} &\text{ if } S=\{s_1,\ldots,s_k\} \text{ and } s_1<\ldots <s_k\end{cases}$$

Our claim is that $f$ is injective. To check injectivity let $S_1,S_2\in 2^\mathbb{R}$, $S_1\neq S_2$:

  • If $S_1,S_2$ are infinite then $f(S_1)=S_1\neq S_2=f(S_2)$.
  • If $S_1,S_2$ have a distinct number of elements then $f(S_1)$ and $f(S_2)$ also do, so $f(S_1)\neq f(S_2)$
  • If $S_1,S_2$ have both $k$ elements, let $S_1=\{a_1,\ldots,a_k\}$ and $S_2=\{b_1,\ldots,b_k\}$ with $a_1<\ldots < a_k$ and $b_1<\ldots < b_k$. Because $S_1\neq S_2$ we can choose $a_i\in S_1\setminus S_2$. If $a_i=a_k$ then $b_k\neq a_k$. If $b_k<a_k$, we have that $\max f(S_1)=2\cdot a_k>2\cdot b_k=\max f(S_2)$, hence $f(S_1)\neq f(S_2)$. If $b_k>a_k$ we also have that $f(S_1)\neq f(S_2)$ with the same argument. In other case, $a_i\neq a_k$. If $a_i= 2· b_k$, then $2·a_k\geq a_k>a_i=2·b_k$ and again $\max f(S_1)>\max f(S_2)$. If $a_i\neq 2· b_k$ then $a_i\in S_1\setminus(S_2\cup\{2 · b_k\})=S_1\setminus f(S_2)\subset f(S_1)\setminus f(S_2)$, so $f(S_1)\neq f(S_2)$.

In any of the cases, $f(S_1)\neq f(S_2)$, so $f$ is injective, which proves that $|2^\mathbb{R}|\leq|2^\mathbb{R}\setminus\{\{x\}:x\in\mathbb{R}\}|$, and so $2^\mathfrak{c}-\mathfrak{c}=2^\mathfrak{c}$.

R.V.N.
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  • It's worth adding that you're using Cantor-Bernstein at the end, to get a bijection from a pair of injections - and that this, while not entirely constructive in general, does not require the axiom of choice. – Noah Schweber Feb 24 '21 at 16:11
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    Taking Cantor-Bernstein as given, I think there's a much simpler proof of the equality $2^\mathfrak{c}-\mathfrak{c}=2^\mathfrak{c}$ (or more fairly, $2^\mathfrak{c}=2^\mathfrak{c}+\mathfrak{c}$ - there is no "cardinal subtraction," really). Specifically, pick some injection $i:\mathbb{R}\rightarrow(0,1)$ and consider the map $$\hat{i}: 2^\mathbb{R}\rightarrow 2^\mathbb{R}: S\mapsto{i(s): s\in S}.$$ This map is injective (since $i$ is) and misses every element of $2^\mathbb{R}$ containing anything outside of $(0,1)$, and there are continuum-many of these (e.g. take ${{x}: x\in [2,3]}$). – Noah Schweber Feb 24 '21 at 16:14
  • You are right, I just meant an easy proof for this that doesn't rely on more general cardinal arithmetic results, so constructive is probably a bad word to define that. And your idea is much better than mine. Choosing ${{x}:x\in[2,3]}$ really simplifies a lot the problems I had because of choosing ${{x} :x\in\mathbb{R}}$. – R.V.N. Feb 24 '21 at 16:17
  • (BTW I don't mean any of the above negatively! I've upvoted this and your other answer.) – Noah Schweber Feb 24 '21 at 16:17
  • If there is no cardinal subtraction, how would $\left|\mathcal{P}(P)\backslash L\right|$ be defined for all points $P$ (where $\mathcal{P}$ denotes the power set)? More specifically, how can $\left|\mathcal{P}(P)\backslash L\right|=\left|\mathcal{P}(P)\right|+\left|L\right|$? – UNOwen Feb 25 '21 at 02:04
  • Cardinal substraction is not defined because it can't be consitently defined. Given two cardinal numbers $\mathfrak{a},\mathfrak{b}$, we can't assign a cardinal number $\kappa$ such that $\mathfrak{a}-\mathfrak{b}=|A\setminus B|=\kappa$ for all sets $A$ and $B$ such that $|A|=\mathfrak{a}$ and $|B|=\mathfrak{b}$. – R.V.N. Feb 25 '21 at 14:17
  • For example, what would you say $\aleph_0 -\aleph_0=|A\setminus B|$ with $|A|=|B|=\aleph_0$ is? If we for example choose $A=\mathbb{N}$, the value of $\aleph_0-\aleph_0$ would completely depend on the choice of $B$. For $B={2n:n\in\mathbb{N}}$ we would have $|A\setminus B|=|{2n+1:n\in\mathbb{N}}|=|\mathbb{N}|$. Instead of that, if we choose $B=\mathbb{N}$, we get $|A\setminus B|=0$. If we choose $B=\mathbb{N}\setminus{2}$, then $|A\setminus B|=1$. – R.V.N. Feb 25 '21 at 14:19
  • This doesn't mean that we can't assign a cardinal number to $|A\setminus B|$ for any sets $A$ and $B$. We only can't assign a value to cardinal substraction, because it would have to work for every couple of sets of a certain cardinality, and it doesn't. – R.V.N. Feb 25 '21 at 14:21
  • Also, in this case we can really asign a cardinal number to $2^\mathfrak{c}-\mathfrak{c}$ because it always holds that if $A$ and $B$ are sets such that $|B|<|A|$, then $|A\setminus B|=|A|$, but it's not really correct to say that that $|A|-|B|=|A|$ because we can't always define that operation. – R.V.N. Feb 25 '21 at 14:24