I would appreciate an explanation of what this means, where $\alpha\in \lambda$, and $\lambda$ is an infinite cardinal.
$(\alpha^{+}\times \alpha^{+})\setminus \{(\alpha,\alpha)\}$
I know $\alpha^{+}= \alpha\cup \{\alpha\}$
To put it in context, this is from part of a proof looking at an initial segment of $\lambda\times \lambda$, as stated: cut out by $(\alpha,\alpha)$, which is the the set $(\alpha^{+}\times \alpha^{+})\setminus \{(\alpha,\alpha)\}$.
It then says thus the initial segment has size at most $|\alpha^{+}|^2$. (Subsequently it says $\alpha$ is an ordinal with $\alpha\lt\lambda$.)
This is especially confusing since an initial segment of a set $S$, say $S_x$ is equal to $\{y\in S: y\lt x\}$, with a strict inequality. So how does $\alpha^{+}$ come into the picture?
Thanks
The notation $(\alpha^{+}\times \alpha^{+})\setminus \{(\alpha,\alpha)\}$ just means exactly what it looks like: $\alpha^+\times \alpha^+$ is the Cartesian product $\{(x,y):x,y\in\alpha^+\}$. In particular, one element of this Cartesian product is $(\alpha,\alpha)$, and so $(\alpha^{+}\times \alpha^{+})\setminus \{(\alpha,\alpha)\}$ is the set of all elements of $\alpha^+\times\alpha^+$ except for $(\alpha,\alpha)$.
As for calling this an initial segment, presumably $(\alpha^{+}\times \alpha^{+})\setminus \{(\alpha,\alpha)\}$ is the set of all elements of $\lambda\times \lambda$ which are less than $(\alpha,\alpha)$, for some ordering of $\lambda\times\lambda$ which would be clear from context. The reason you would use $\alpha^+$ rather than $\alpha$ is that pairs like $(\alpha,\beta)$ or $(\beta,\alpha)$ for $\beta<\alpha$ are less than $(\alpha,\alpha)$. If you just took $\alpha\times\alpha$, that set would not contain such pairs.