Is there a function $f : N\to N$ such that every $(k-1)$-connected graph with minimum degree at least $f(k)$ is at least $k$-connected?

Question

I tried to solve the following problem:

Is there a function $f: N \to N$ such that every $(k -1)$-connected graph with minimum degree, at least $f(k)$ is at least $k$-connected?

I have understood what k-connectivity means and also the meaning of the minimal degree. But I don't get to connect the two things. Does anyone have any tips on the solution for a function?

Thank you very much!

You can learn mathjax very easily here. https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference/29979#29979 Try to use it next time. — Sumanta, Sep 12 '20 at 09:52
A $0$-connected (i..e, arbitary) graph can fail to be $1$-connected, no matter what we know about its minimal degree. — Hagen von Eitzen, Sep 12 '20 at 09:55
https://math.stackexchange.com/questions/3388974/find-a-graph-such-that-kappag-lambdag-deltag/3389232#3389232 — bof, Sep 12 '20 at 12:37

score 3 · Answer 1 · answered Sep 12 '20 at 10:02

3

Fix $m$ and $n$ with $n\gg m$ and consider a "$K_m$ made of $K_n$'s", i.e., we have $m$ disjoint copies of $K_n$'s, pick a vertex from each of these, and add edges between any two of the $m$ picked vertices. The minimal degree of this graph is $n$, but the connectivity is determined by $m$.

answered Sep 12 '20 at 10:02

Hagen von Eitzen

382,479

Thanks for reply! How do you conclude from this that there is no function? – Lea Blecher Sep 17 '20 at 17:15

Is there a function $f : N\to N$ such that every $(k-1)$-connected graph with minimum degree at least $f(k)$ is at least $k$-connected?

1 Answers1