Prove or disprove - order types and constants

Question

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?

Answer 1

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 $2014+\omega=\sup\{2014+n\mid n\in\omega\}$.