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..


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.

  • 2
    $\begingroup$ Hint: subspaces are closed subgroups. $\endgroup$
    – martini
    Commented Jul 9, 2014 at 8:34
  • 1
    $\begingroup$ For the singletons case just note that any subgroup must contain $0$, and so can only contain $0$. $\endgroup$ Commented Jul 9, 2014 at 8:37
  • $\begingroup$ @martini : Ok, So, copies of $\mathbb{R}^k$ are closed connected subgroups for $k<n$.. Can i get any other subgroups.. $\endgroup$
    – user87543
    Commented Jul 9, 2014 at 8:37
  • $\begingroup$ @HennoBrandsma : Yes Yes.. i got it :) $\endgroup$
    – user87543
    Commented Jul 9, 2014 at 8:39
  • $\begingroup$ Martini gave the hint: linear subspaces. Like the line $x=y$ in the plane, etc. And strictly speaking, $\mathbb{R}$ is not even a subset of $\mathbb{R}^2$... $\endgroup$ Commented Jul 9, 2014 at 8:39

1 Answer 1


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.


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