Is $\aleph_1 - \aleph_0$ also indeterminate?
closed as unclear what you're asking by user99914, Stefan4024, Ove Ahlman, Claude Leibovici, Rebellos Nov 29 '17 at 10:53
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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 $\beta - \alpha = \beta$.