Definition of $\mathfrak a$-connected graphs?

Question

A typical definition of a graph $G$ being $k$-connected ($k\in\mathbb N_0$) is this:

$k<|G|$ (the order of $G$) and for $X$ being a subset of $V(G)$ such that $|X|<k$ holds $G\setminus X$ is connected.

I would like to extend this definition to $\mathfrak a$-connectivity of infinite graphs. There are two likely definitions that would work:

$\mathfrak a<|G|$ (the order of $G$) and for $X$ being a subset of $V(G)$ such that $|X|<\mathfrak a$ holds $G\setminus X$ is connected.

$\mathfrak a + 1 \leq |G|$ (the order of $G$) and for $X$ being a subset of $V(G)$ such that $|X|<\mathfrak a$ holds $G\setminus X$ is connected.

Since the addition in the second definition is cardinal addition, both definitions simplify to the finite case. But if $\mathfrak a$ is infinite, the first condition in the second definition becomes $\mathfrak a \leq |G|$.

My main reasoning is this: the simple complete bipartite graph $K_{n,n}$ is $n$-connected. So the graph $K_{\omega,\omega}$ should be $\omega$-connected. The second definition allows this, the first definition does not. Nevertheless, in the first definition we would still have $$\sup\{k\in\mathbb N_0:K_{\omega,\omega}\mbox{ is }k\mbox{-connected}\} = \omega$$

Question: Are there precedents in the literature for either generalized definition?

If you don't know a reference, but a good reason to prefer one definition over another, leave a comment!

Answer 1

For finite graphs, the second part of the definition is the important one and we usually don't think about the first part at all.

The reason that the first part exists is complete graphs: $K_n$ has no set $X$ such that $K_n - X$ is not connected (aside from some sophistry about the $0$-vertex graph), so if we didn't have the first part, then $K_n$ would be $k$-connected for all $k$. We don't want that, because $K_n$ doesn't share most of the properties of $k$-connected graphs for $k \ge n$; notably, its minimum degree is only $n-1$.

So we make a separate definition that could be phrased the way you've done it, but effectively says "For $K_n$, ignore the usual definition; this graph is $(n-1)$-connected but not $n$-connected."

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Similarly, for infinite graphs, we should try to go as far as we can with the important part of the definition. $K_{\omega,\omega}$ should be $\aleph_0$-connected, because for any $X$ with $|X| < \aleph_0$, $K_{\omega,\omega}-X$ is still connected.

Once again, the only graphs we have to have an extra clause for are complete graphs. It seems perfectly reasonable for the complete graph $K_\omega$ to be $\aleph_0$-connected, but of course it definitely shouldn't be $k$-connected for any $k > \aleph_0$. So add whatever clause you need that ensures this; the second definition seems like it works fine.

You can also generalize any of the other definitions of $k$-connectivity for finite $k$. For example, the definition in West's Introduction to Graph Theory is

The connectivity of $G$, written $\kappa(G)$, is the minimum size of a vertex set $S$ such that $G-S$ is disconnected or has only one vertex. A graph is $k$-connected if its connectivity is at least $k$.

This survives intact when our graphs are infinite.