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Given $\alpha \in Ord$ and $\gamma \in Lim$

where $Ord$ is the class of all ordinal numbers and $Lim$ class of all limit ordinals

(i.e. those which are not succesors),

I was wondering how to prove that in that case $\alpha$ + $\gamma$ $\in Lim$ too.

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    What is your definition of $\alpha+\gamma$? Can you show it has no largest element? – 2017-02-17
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    The definition is $\alpha + \gamma = \cup(\alpha+\delta)$ where union goes over all $\delta<\gamma$ – 2017-02-17
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    Alright, so suppose $\alpha+\gamma$ were not a successor. It would have some last element $\eta$. This $\eta$ would be in $\alpha+\delta$ for some $\delta$, by definition. Do you see how to get a contradiction now? *HINT: is $\delta$ the biggest element of $\gamma$? If not, what happens when we look at e.g. $\alpha+(\delta+1)$?* – 2017-02-17

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Note that an ordinal, as a linear order, is a limit ordinal if and only if it does not have a last element.

Next, note that a linear order does not have a last element if and only if there is a tail segment without a last element.

Finally, the ordinal sum $\alpha+\gamma$ can be defined as an ordinal with an initial segment of order type $\alpha$ and a tail segment with order type $\gamma$.

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    I have one direct way of proving. Namely, assume $\beta < \alpha + \gamma$. Hence, there is $\delta < \gamma$ such that $\beta < \alpha + \delta$. This implies $\beta + 1 < \alpha + (\delta + 1)< \alpha + \gamma$. But I don't get why there exists $\delta < \gamma$ such that $\beta < \alpha + \delta$ – 2017-02-17
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    If $\beta<\alpha$, then obviously so; otherwise take the order type of $\beta\setminus\alpha$. – 2017-02-17
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    What exactly do you mean by taking the order type of $\beta$ \ $\alpha$? – 2017-02-17
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    $\beta\setminus\alpha$ is a set of ordinals. Then there is some ordinal to which it is isomorphic. Show that this ordinal has to be less than $\gamma$. – 2017-02-17