Let $K $ be the connected component of a lie group $G$ containing identity. Then why is it an open set?
I thought that any connected component of a manifold is open because it is locally homeomorphic to $\mathbb{R^n}$ (i.e around every point) but another reasoning that I saw elsewhere is that it is due to local connectedness of $G$. What does that mean?
Also my reasoning won't work for manifolds with boundary, so is any modification of it applicable or is the result itself not true if we consider the manifolds with boundary?
The reasoning is coming from Spivak's Differential Geometry Vol 1, Chapter 10, p-373.
If $G$ is a topological group, the identity component is open if and only if $G$ is locally connected. Note that for a topological group, "locally connected" means that there's a connected open neighborhood of the identity; that this is true for all other points follows by multiplying this neighborhood by $g$. This is in general true for locally whatever: just do it at the identity.
If the identity component is open, this is of course a connected open neighborhood of the identity. Conversely, if there's a connected open neighborhood of the identity $U$, then because the identity component $G_e$ is a subgroup, $gU \subset G_e$ whenever $g \in G_e$. So we may write $G_e = \cup_{g \in G_e} gU$, where the right hand side is a union of open sets.
An explicit case of a non-locally-connected group, where this fails, is $\Bbb Q$.