Infinite set of graphs neither of which are homeomorphic

Question

Show that there is an infinite set of graphs $\{G_1,G_2,...\}$, where:

$\quad \quad \quad$ $\forall i, \forall j, i\neq j :$ graph $G_i$ isn't homeomorphic to graph $G_j$

I have a hard time figuring this out. Any help is appreciated.

Answer 1

I couldn't comment this (not enough reputation) so here it is as an answer:

Can you think of an infinite set of graphs? Maybe by varying the number of vertices?

Can you show that if, say, a pair of (a certain kind of) graphs has a different number of vertices or edges, that you can't have a homeomorphism between them?

Answer 2

A simple example is to have $G_i$ be a graph with $i$ vertices and no edges. A more involved example would be subdivisions of a graph with two vertices and $i$ edges between them (which are subdivided so that they qualify as simple graphs).

Answer 3

Start with $G_0$ let $G_0$ be the empty graph on $\aleph_0$ many vertices.
$G_1$ has $\aleph_0$ many vertices and $1$ edge.

$G_2$ has $\aleph_0$ many vertices and $2$ edges, such that these edges are non-adjacent.

Assume that for all $i<k$, $G_i$ has been defined.
$G_k$ has $\aleph_0$ many vertices and $k$ edges such that no two distinct edges share a common end.

For $i\ne j$, $G_i$ is not homeomorphic to $G_j$, because they have a different number of connected components.

Answer 4

For each $n\ge 2$: a central vertice and $n$ vertices united to the central vertice by a edge each one.