I am confused by how these facts are consistent:
i. If $A$ and $B$ are disjoint sets, then $|A| + |B| = |A \cup B|$ [1]
ii. If $S = \{x : x \in [0,1] \} $, then $|S| \neq 0$, $|\mathbb{R} – S| = |\mathbb{R}|$, and $ |\mathbb{R} – S| + |S| = |\mathbb{R} | $
iii. If $a + b = a$, then $b=0$ [2]
Given ii. $|\mathbb{R} – S| + |S| = |\mathbb{R} – S |$, in which case by iii. $|S| = 0$, contradicting ii. But ii. seems to follow from i. and the observations that the cardinality of the set of points along the real number line is equal to the cardinality of the set of points in any segment of the real number line, ($S$ and $\mathbb{R} – S $ being two such segments) [3] and that $S$ and $\mathbb{R} – S $ are disjoint.
I'm tempted to say that what we learn is that i. is false and so that the function $f: X \mapsto |X| $ is not $\sigma $-additive. Alternatively, perhaps iii. is not true for ''$+$'' in i. But I think then I struggle to understand what ''$+$'' actually means in this context.
Am I right? Or have I gone wrong somewhere?
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Best Answer
It's 'iii' that is not true. For cardinal arithmetic, if $|A|=a$ and $|B|=b$, then $a+b$ is defined to be the cardinality of the disjoint union of $A$ and $B$ — for instance, it's the cardinality of $\{\langle a,0\rangle:a\in A\}\cup\{\langle b,1\rangle: b\in B\}$. You can convince yourself that this corresponds to the usual addition in the case where $a$ and $b$ are finite, but we have e.g. $\aleph_0+\aleph_0=\aleph_0$ and in fact, $\mathfrak{a}+\mathfrak{b}=\max(\mathfrak{a},\mathfrak{b})$ if either of $\mathfrak{a}$ or $\mathfrak{b}$ is infinite. This is distinct from ordinal addition, which is defined by appending one order to another; this also gives the usual result for finite numbers, but for instance $\omega+\omega$ isn't equal to $\omega$. ($\omega+\omega$ has a member with no immediate predecessor, but $\omega$ doesn't.)
Incidentally, cardinal multiplication is defined similarly; $a\times b$ is the cardinality of the cartesian product $A\times B=\{\langle a,b\rangle:a\in A, b\in B\}$, and this definition gives the usual result for finite $a$ and $b$, but again we have $\mathfrak{a}\times\mathfrak{b}=\max(\mathfrak{a},\mathfrak{b})$ if either $\mathfrak{a}$ or $\mathfrak{b}$ is infinite.