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Let $k$ and $l$ be natural numbers and let $\omega=[(\mathbb N, \le)], \ \eta=[(\mathbb Q, \le)]$ be order types (or ordinals).

Prove or disprove the following:

  1. if $k+\eta=l+\eta$ then $k=l$.

  2. if $k+\omega=l+\omega$ then $k=l$.

  3. $2014\cdot\omega=2014+\omega$

For 1, I'm not sure, the only way for $k+\eta=l+\eta$ to be true is if the two constants are equal, because if they weren't then it wouldn't be true. That's obviously not formal enough.

For 2, we know from ordinal arithmetic that $n\in\mathbb N+\omega=\omega$ so let's take two different natural numbers instead of $k$ and $l$ and we'll get that the first term is true but the second isn't. Is that enough to prove it ?

For 3, this is the way I see it: $\{2014,2\cdot 2014,...n\cdot2014\}=2014\cdot\omega\neq2014+\omega=\omega$, is it enough?

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HINTS:

  1. You can prove this by induction on $k$.
  2. Yes, this is enough to disprove this. For concreteness, $k=0, l=1$.
  3. Recall that $2014\cdot\omega=\sup\{2014\cdot n\mid n\in\omega\}$, and that $2014+\omega=\sup\{2014+n\mid n\in\omega\}$.
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  • $\begingroup$ So 1 is true ? what is the base of induction ? About 3, why are they equal to these supermums ? $\endgroup$ – GinKin Mar 10 '14 at 14:12
  • $\begingroup$ Well, for $k=0$ note that if $\eta=l+\eta$, then $l=0$, otherwise one of the orders has a least element and the other doesn't. As for the third, how do you define the addition and multiplication of two well-orders? $\endgroup$ – Asaf Karagila Mar 10 '14 at 14:13
  • $\begingroup$ Our professor told us the it goes like this $\{what\}\cdot\{how \ much\}$ so for example: $\omega\cdot 2014=\{\omega\cdot\omega\cdot...\cdot\omega\}$ 2014 times. $\endgroup$ – GinKin Mar 10 '14 at 14:17
  • $\begingroup$ Are you sure that you didn't misunderstand your professor? $\endgroup$ – Asaf Karagila Mar 10 '14 at 14:18
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    $\begingroup$ Well, you might also have heavily misunderstood (or been misexplained by your teacher). "What" and "How much" are informal ways to say that we replace each point in the right order with a copy of the left order (so $\omega\cdot 2$ is to replace each point in $2$ by a copy of $\omega$, so it's $\omega+\omega$). Then you can show that $2014\cdot\omega$ is isomorphic to $\omega$. And no, this (nor the third) is not true for every infinite ordinal. $\endgroup$ – Asaf Karagila Mar 10 '14 at 15:08

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