I'm reading A Course in Topological Combinatorics, written by Longueville (Universitext, Springer, 2012).
In a topological proof of Theorem 2.2, following:
Greene, J.E., A new short proof of Kneser’s conjecture, Amer. Math. Monthly 109, 918-920, 2002.
is said:
$$U_i=\{x\in \mathbb{S}^d\colon \text{there exists a $k$-set $S \subset X$, $c(S)=i$, $S\subset H(x)$} \}$$ is open. Where $$H(x)=\{y\in \mathbb{S}^d \colon \langle x,y\rangle>0\}.$$ is an open hemisphere with $x$ as pole. And a $k$-set is a subset of $X$ with $k$ elements.
In order to prove the set is open I have come up with an argument involving continuous group actions, $SO(d+1)$ acting on $\mathbb{S}^d$ to be concrete.
And it would work if the following assertion is true:
If $\widetilde{G}$ is an open neighborhood of the identity in $SO(d+1)$, and $x\in \mathbb{S}^d$, then $$\widetilde{G}\cdot x:=\bigcup_{g\in \widetilde{G}} g \cdot x$$ is open.
If my intuition is not betraying me, this should be open (and if you think about it in low dimensions it seems true) and it should be an obvious consequence of the fact that the action is continuous and transitive. However I don't come up with proof. So is it just obvious and I don't need to prove anything or I am blundering.
Any help or hint would be appreciated.
Your assertion is true. Let me state it in more general terms. Let $G$ be a compact topological group acting continuously and transitively on a Hausdorff space $X$ and let $z \in X$. If $U$ is an open neighbourhood of the identity, then $U \cdot z$ is open in $X$.
We can prove this in two steps. First consider the isotropy subgroup of $z$
$$ G_z = \{ g \in G \mid g \cdot z = z \} $$
If you put the quotient topology on $G/G_z$, then the map $G \to G/G_z$ that sends $g$ to $gG_z$ is continuous and open.
Now consider the map $G/G_z \to X$ that sends $gG_z$ to $g \cdot z$. You can prove that this map is a homeomorphism (here is where you use that $G$ is compact and $X$ is Hausdorff). In particular, it is open.
Then the composition of these two maps $G \to X$ is open and it sends $U$ to $U \cdot z$, so $U \cdot z$ is open. I hope this helps, I can expand on the steps if you wish.