Question is :
What are closed connected subgroups of $\mathbb{R}$ and from that deduce what are closed connected subgroups of $\mathbb{R}^n$
What i have done so far is :
Only connected subsets of $\mathbb{R}$ are singletons and intervals..
As i need my connected sets to be closed i need closed intervals (singletons are already closed)
Choose $A=\{a\}$ for $a\neq 0$...
If we want $A$ to be subgroup then we need $\{na : n\in \mathbb{Z}\}\subset A$ which is not possible
Thus, only closed connected singleton is $\{0\}$
Now choose $A=[a,b]\subset \mathbb{R}$.. As $b\in [a,b]$ and if we want this to be subgroup we need $\{nb : n\in \mathbb{Z}\}\subset A$ which is not possible.
Thus there is no bounded closed connected subgroup...
So, only closed connected subgroups of $\mathbb{R}$ are $\{0\}$ and $\mathbb{R}$
I am not very sure about how should i deduce about closed connected subgroups of $\mathbb{R}^n$..
I somehow guess that only such subgroups are $\{0\}$ and $\mathbb{R}^n$ but i am not sure..
I would be thankful if some one can give some hints to workout this..
EDIT :
I have tried to show that this is indeed a subspace..
For $H$ a closed connected subgroup of $\mathbb{R}^n$ Suppose $x,y\in H $ then as $H$ is a subgroup $ax\in H$ and $by\in H$
Then i immediately wanted to write $ax+by\in H$ (hoping that closed and connected implies path connected ) which would imply $H$ is a subspace and everything would go so smoothly as there are only finitely many subspaces..
But then i realized closed and connected does not imply path connectedness...
I do not know how to proceed.
Perhaps this is an overkill, but you can prove this with Lie theory: The Lie algebra of $\mathbb{R}^n$ is $\mathbb{R}^n$, and the exponent is the identity. If $G$ is a closed connected subgroup then its Lie algebra $\mathfrak{g}$ is a subspace of $\mathbb{R}^n$ and $\exp\mathfrak{g}=\mathfrak{g}$ generates $G$. But $\mathfrak{g}$ is already a subgroup of $\mathbb{R}^n$, so $G=\mathfrak{g}$ and it is a subsapce.