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Intuitively, it seems to me that $|\mathbb{R}^{2}| > |\mathbb{R}|$. My intuition says something to the effect of: There exists a bijection between $\mathbb{R}$ and $\{(x,0) : x \in \mathbb{R}\}$, and the second set is a subset of $\mathbb{R}^{2}$.

But I remember reading somewhere that $|\mathbb{R}^{2}| > |\mathbb{R}|$ is not true, though I may be mistaken.

I know only the very basics about countable sets which are infinite. I also know that $\mathbb{R}$ is uncountable.

To be clear, I am looking for:

  • Is $|\mathbb{R}^{2}| > |\mathbb{R}|$?
  • Why?
  • Any quick resource/lecture notes to quickly brush up on related concepts.

The most relevant question I could find was: Is it proper to say that two infinite sets are the "same size" if there is a bijection between them?

David Sainez
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    That there exists a bijection from $\Bbb R$ to a proper subset of $\Bbb R^2$ does NOT mean $|\Bbb R^2|>|\Bbb R|$. The definition of $|A|=|B|$ is there exists a bijection between $A$ and $B$, and it is extremely likely that $|A|=|A'|$ where $A'\subsetneq A$, say $|\Bbb Q|=|\Bbb Z|=|\Bbb N|$. – Vim Jul 20 '17 at 04:48
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    Indeed, the sets have the same cardinal number. One can set up a bijection in numerous ways, it just cannot be continuous. – infinitylord Jul 20 '17 at 05:07
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    As an aside, although not true for finite sets, you do have for any infinite set $A$ that $|A|=|A\times A|=|A\times A\times A|=\dots = |\underbrace{A\times\cdots \times A}_{n~\text{times}}|$ for any finite value of $n$. – JMoravitz Jul 20 '17 at 05:31

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