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Let's say we have $\omega + \omega$. Since these sets are not disjoint we can replace them by disjoint sets of the same cardinality, namely $\omega \times \{0\}$ and $\omega \times \{1\}$.

Then $\big | \omega \times \{0\} \big| + \big | \omega \times \{1\}\big | = \big | \omega \times \{0\} \cup \omega \times \{1\}\big |$.

I really do not understand what the point of this is since I feel like the end result would still be $\omega + \omega$.

I would appreciate a concrete example of cardinal arithmetic.

Asaf Karagila
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user7090
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  • But what are you trying to prove ? The fact that, in order to "sum" the cardinalities, we have that the sets must be disjoint is easy to grasp; if $A = { 1, 2, 3 }$ (with $|A| = 3$) and $B = { 2, 4 }$ (with $|B| = 2$), we have that : $A \cup B = { 1, 2, 3, 4 }$ (with $|A \cup B| = 4$). So $|A \cup B| \ne |A| + |B|$. – Mauro ALLEGRANZA Apr 06 '14 at 18:37
  • See this post and in Wiki Cardinal number : Cardinal addition If $X$ and $Y$ are disjoint, addition is given by the union of $X$ and $Y$. If the two sets are not already disjoint, then they can be replaced by disjoint sets of the same cardinality, e.g., replace $X$ by $X \times { 0 }$ and $Y$ by $Y \times { 1 }$. – Mauro ALLEGRANZA Apr 06 '14 at 19:14

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Let me write $\aleph_0$ instead of $\omega$ because it makes sure there is no confusion between ordinal and cardinal arithmetic.

Your calculation is correct. $\aleph_0+\aleph_0=\aleph_0+\aleph_0$. But note that $\aleph_0+\aleph_0$ is not a cardinal itself. The point of cardinal arithmetic is to try and simplify the result so you will have a cardinal. Sometimes we have to involve exponentiation, in which case it is sometimes have to remain "open" (e.g. $2^{\aleph_0}\cdot\aleph_2$ cannot be simplified further without assumptions on the continuum); but at least for addition we want, and can, simplify the result as much as possible.

The point is that $\aleph_0+\aleph_0=\aleph_0$. So $|\omega|+|\omega|=|\omega|$. To prove this you need to show that there is a bijection between $\omega\times\{0\}\cup\omega\times\{1\}=\omega\times2$ and $\omega$ itself. This has to be done "at the low level", at least for the first couple of times.

But after you've proved all sort of basic theorems about cardinal arithmetics, then you can go ahead and prove something of the following flavor: $|\Bbb R^\Bbb N|=|\Bbb R|$:

$$\left(2^{\aleph_0}\right)^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}$$

And all that without using a single explicit bijection.

Here is another example (taken from my answer here).

Suppose that $E$ is an equivalence relation on $\Bbb R$ such that every equivalence class is (at most) countable, then $|\Bbb R/E|=|\Bbb R|$ (that is, there are $2^{\aleph_0}$ equivalence classes).

Let $\{R_i\mid i\in I\}$ be the set $\Bbb R/E$ (where $I$ is such index set that each $R_i$ is distinct). Then we have: $$2^{\aleph_0}=|\Bbb R|=|\bigcup\{R_i\mid i\in I\}|=\sum_{i\in I}|R_i|=|I|\cdot\sup\{|R_i|\mid i\in I\}=|I|\cdot\aleph_0=|I|$$

(We assume here that there exists an infinite equivalence class, or at least that the sizes of the equivalence classes are unboundedly finite; if they are, replace the last $\aleph_0$ by their upper bound, the result is the same.)

And all that without appealing to concrete bijections, injections or surjections. (Although we do appeal to the axiom of choice, twice.)

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