# Is $\infty - \infty$ still indeterminate when the infinities are of different cardinality? [closed]

Is $\aleph_1 - \aleph_0$ also indeterminate?

## closed as unclear what you're asking by user99914, Stefan4024, Ove Ahlman, Claude Leibovici, RebellosNov 29 '17 at 10:53

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• What's $\aleph_1-\aleph_0$ mean? – user223391 Nov 29 '17 at 0:59
• math.stackexchange.com/questions/140930/… – user223391 Nov 29 '17 at 1:01
• @Jake what do you see going on that would count as harassment? One useful tool of teaching is in asking leading questions. Generally through the use of asking questions, the one being asked can answer and see how that question and answer relates to their original question. – JMoravitz Nov 29 '17 at 1:10

These are completely unrelated questions.

The notion of extended real number — the kind of thing which $\infty$ and $-\infty$ are — is geometric in nature; it has absolutely nothing to do with cardinality.

The reason $\aleph_0 - \aleph_0$ is undefined in cardinal number arithmetic is completely different than the reason $\infty - \infty$ is undefined in extended real number arithmetic.

In cardinal number arithmetic, it would be reasonable to define $\aleph_1 - \aleph_0 = \aleph_1$. Or more generally, if $\alpha$ and $\beta$ are cardinal numbers with $\alpha < \beta$ and $\beta$ infinite, then it would be reasonable to define $\beta - \alpha = \beta$.