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I answered the first part as follows:

$\omega + 1 = \{0, 1, 2, 3, 4, 5, ..., \omega\} = \aleph_0 $ Since every element of ω is finite and one element in $\omega + 1$
is infinite, they are clearly different infinite numbers. This is also known as an ordinal number. Hence, $\omega +1$ is countable but it doesn't represent a higher infinity. It is infinity itself.

But when it comes to $\omega + \omega$, I'm a bit confused on how to approach it.

Becky445
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  • You have the bijection $\omega+1\to\omega$ given by $\infty\mapsto0, n\mapsto n+1$ for $n\in\omega$. –  Feb 29 '20 at 21:11
  • For $\omega+\omega\to\omega$ send $n\to 2n$, for $n\in$ the first $\omega$ and $n\mapsto 2n+1$ for $n$ in the second $\omega$ of $\omega+\omega$. –  Feb 29 '20 at 21:13
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    @flan You should make that an answer. – Noah Schweber Feb 29 '20 at 21:15

1 Answers1

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$\omega+1$ is an order type (it represents a countable well-order with a unique maximum beyond a copy of $\omega$ while $\omega+\omega$ is two copies of $\omega$ ordered one left to the other. Both represent countable sets in cardinality, so they're not a higher order of infinity.

The set of all countable well-orders is of a higher cardinality, called $\omega_1$ (as a well-order) and $\aleph_1$ (as a cardinal number).

Henno Brandsma
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