Another ordinals question $\sum_{i\in\omega +3} i\cdot\omega + \omega\cdot i$

Question

So I have to calculate sum $$ \sum_{i\in\omega +3} i\cdot\omega + \omega\cdot i$$ If I got it right, $1\cdot\omega+2\cdot\omega+\dots+(\omega-1)\cdot\omega = \omega\cdot(\omega-1)$ Then I have to calculate this: $\omega\cdot1+\omega\cdot2+\dots+\omega\cdot(\omega-1)+\omega\cdot(\omega-1)$ And I don't have idea how to approach it. If I go by natural numbers formula i get division by 2 and I cant find that case in ordinal arithmetics.

Answer 1

By associativity, we have:

\begin{align} \sum_{i\in\omega+3}i\cdot\omega+\omega\cdot i\quad=\quad &\left(\sum_{i\in\omega}i\cdot\omega+\omega\cdot i\right)\\ &+\quad (\omega\cdot\omega+\omega\cdot\omega)\\ &+\quad((\omega+1)\cdot\omega+\omega\cdot(\omega+1))\\ &+\quad((\omega+2)\cdot\omega+\omega\cdot(\omega+2)) \end{align}

Some of these terms are easy, so I hope you agree with the following: \begin{align} &\omega\cdot\omega+\omega\cdot\omega=\omega^2+\omega^2=\omega^2\cdot 2\\ &\omega\cdot(\omega+1)=\omega^2+\omega\\ &\omega\cdot(\omega+2)=\omega\cdot(\omega+1)+\omega=\omega^2+\omega+\omega=\omega^2+\omega\cdot 2 \end{align} For the rest we need to reason with limits. For example $(\omega+1)\cdot \omega$ is the supremum of the set $\{(\omega+1)\cdot\gamma\mid\gamma\in\omega\}$. You can prove that $(\omega+1)\cdot\gamma=\omega\cdot\gamma+1$ for any $\gamma\geq1$:

If $\gamma=1$ we have $(\omega+1)\cdot 1=\omega+1$

If $\gamma=\beta+1$, we have $(\omega+1)\cdot(\beta+1)=(\omega+1)\cdot\beta+(\omega+1)$. Using the induction hypothesis this equals $(\omega\cdot\beta+1)+(\omega+1)=\omega\cdot\beta+(1+\omega)+1=\omega\cdot\beta+\omega+1=\omega\cdot(\beta+1)+1$.

Hence we see that $(\omega+1)\cdot\omega$ is the supremum of $\{\omega\cdot\gamma +1\mid\gamma\in\omega\}$. You can show that $\omega\cdot\gamma+1$ is unbounded in $\omega^2$ (hint: if $\alpha\in\omega^2$, then $\alpha<\omega\cdot\beta$ for some $\beta\in\omega$), and you can show that $\omega\cdot\gamma+1<\omega^2$ for any $\gamma\in\omega$. Therefore the supremum of this set is $\omega^2$. So we see that $(\omega+1)\cdot\omega=\omega^2$.

In a similar way you can compute $(\omega+2)\cdot\omega$.

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Now all that is left is $\sum_{i\in\omega}i\cdot\omega+\omega\cdot i$. However, this becomes quite easy: $i\cdot\omega=\omega$ for any $i\in\omega$ (I showed how to prove this in my answer to your previous question), and $\omega\cdot i$ is just $\omega\cdot i$. Finally note that $\omega+\omega\cdot i=\omega\cdot(i+1)$ (prove this by induction).

So we have $\sum_{i\in\omega}i\cdot\omega+\omega\cdot i=(\omega\cdot 2)+(\omega\cdot 3)+(\omega\cdot 4)+\dots$. It is not too difficult to show that this is equal to $\omega^2$.

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So, if I made no mistakes, in conclusion we should have: \begin{align} \sum_{i\in\omega+3}i\cdot\omega+\omega\cdot i\quad=\quad &\quad\omega^2\\ &+\quad \omega^2\cdot 2\\ &+\quad\omega^2+\omega^2+\omega\\ &+\quad\omega^2+\omega^2+\omega\cdot 2\\ =\quad&\omega^2\cdot 7+\omega\cdot 2 \end{align}

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Note that we never reason with subtraction, division or reordering addition / multiplication, since these things are invalid in ordinal arithmetic. Instead, we reason with unboundedness, to show suprema are larger than or equal to a given ordinal, and with boundedness to show suprema are smaller than or equal to a given ordinal. This is the way you should think about ordinal addition.

As with all mathematics, the only way to really get comfortable with this is by practice, so I encourage you to try computing some different sums yourself.