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Need guidance on this problem. Given the following ordinals, determine the order relation (and equalities) - ($\Omega = 2^{\aleph_0}$)

$\omega_1$, $\Omega \cdot 3$, $\omega \cdot \omega_1$, $3^{\Omega}$, $\Omega^{\omega_1}$

My solution (partial):

$\Omega\cdot 3 \gt \Omega\cdot 2 \gt \Omega$, $3^{\Omega} = \lim_{n \lt \Omega}3^{n} = \Omega$ ($\Omega$ is a limit ordinal)

I'm not certain about how to deal with $\omega \cdot \omega_1$ and $\Omega^{\omega_1}$

The inequalities:

$\Omega\cdot 3 \gt 3^{\Omega} \gt \omega_1$

Thanks!

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    $\Omega$ is the size of the continuum (cardinal exponentiation). All other exponentiations in this question are in the ordinal sense.2011-02-08

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A hint for $\omega\cdot\omega_1$: What does it look like? Can you simplify it? (For a simpler example, compare $2\cdot\omega$ and $\omega\cdot2$...)

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    Not exactly: $\omega\cdot\omega_1$ is the result of concatenating $\omega_1$ copies of $\omega$, and you should see that this is *precisely* $\omega_1$.2011-02-08