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I was playing with $ \aleph_0 $ on Wolfram Alpha when I encountered this:

I understand why $ \aleph_0 + 1 = \aleph_0 $ but why is $ \aleph_0 - 1 = \text{undefined} $?

Lone Learner
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  • You can define $\aleph_0-1$, since there is a unique cardinal $\kappa$ such that $\kappa+1=\aleph_0$, namely $\aleph_0$. As for why this equality holds, this is the statement that if you have a countably infinite set and add one element, then the resulting set will again be countably infinite. – Wojowu Oct 14 '20 at 16:25
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    @Wojowu Is Wolfram Alpha making an error in saying that $ \aleph_0 - 1 = \text{undefined} $? – Lone Learner Oct 14 '20 at 16:34
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    It's probably not so much an error, but rather an artifact of "subtraction" for cardinals being undefined in general. For example, $\aleph_0-\aleph_0$ really is undefined. If one of the cardinals is finite, like $1$, then you can make a sensible answer. But it's not a very useful operation on the whole (unlike addition which makes sense for all cardinals). – halrankard2 Oct 14 '20 at 16:42
  • @halrankard2 I think you could basically write that as an answer instead of a comment. – Mark S. Oct 14 '20 at 18:38

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Following up on comments:

Given sets $X$ and $Y$, if $\kappa=|X|$ and $\lambda=|Y|$, then we have a well-defined cardinality (or cardinal) $\kappa+\lambda=|(X\times\{0\})\cup (Y\times\{1\})|$. (The $0$ and $1$ factors are present to force the sets to be disjoint, which is important when dealing with finite cardinals.) This gives a well-defined notion of addition of cardinals, which extends the standard definition of addition for finite nonnegative integers.

On the other hand, there is no general notion of "subtraction" for cardinals. For example if I have a set $X$ of cardinality $\aleph_0$, and a subset $Y$ also of cardinality $\aleph_0$, then $X\setminus Y$ could have any cardinality from $0$ to $\aleph_0$. So $\aleph_0-\aleph_0$ has no well-defined value. This is similar to how $\infty-\infty$ is an indeterminate when doing limits.

So this is probably the reason why Wolfram Alpha returns "undefined" when you enter a subtraction involving non-finite cardinals. However, as Wojowu says, there are some cases where subtraction does make sense. For example, if $\kappa$ is infinite ($\aleph_0$ for example) and $n$ is finite, then there is a unique cardinal that "solves" the equation $n+x=\kappa$, namely $\kappa$ itself. So one could reasonably say $\kappa-n=\kappa$. For a general discussion, see: Cardinal number subtraction

halrankard2
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