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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:

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


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To be clear, $\Omega = 2^{\aleph_0}$ is ordinal exponentiation or cardinal exponentiation? – Jason DeVito Feb 8 '11 at 16:16
$\Omega$ is the size of the continuum (cardinal exponentiation). All other exponentiations in this question are in the ordinal sense. – Andrés E. Caicedo Feb 8 '11 at 16:28

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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$\omega \cdot 2 = \omega + \omega$ while $2 \cdot \omega = \omega$, so $\omega_1 \gt \omega \cdot \omega_1 \gt \omega$? Also, am I right about the others? – Ma.H Feb 8 '11 at 17:11
Not exactly: $\omega\cdot\omega_1$ is the result of concatenating $\omega_1$ copies of $\omega$, and you should see that this is precisely $\omega_1$. – Andrés E. Caicedo Feb 8 '11 at 17:36

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