Question

Prove that if \(\mathcal{X}\) is a normed vector space, then for any subspace \(\mathcal{S} \subset \mathcal{X}\), \(\overline{S}\) is a subspace.

Answer 1

I think we are assuming that the topology on the space is the metric topology induced by the norm.

Answer 2

It looks like we have three cases to deal with.

1) Where \(x, y \in \mathcal{S}\), we show that \(ax + by \in \overline{\mathcal{S}}\)
2) Where \(x'\) is a limit point of \(\mathcal{S}\) and \(y \in \mathcal{S}\), we show that \(ax' + by \in \overline{\mathcal{S}}\)
3) Where \(x', y'\) are limit points of \(\mathcal{S}\), we show that \(ax' + by' \in \overline{\mathcal{S}}\)

Answer 3

1) Is obvious since \(\mathcal{S}\) is a subspace and so \(ax + by \in \mathcal{S} \subset \overline{\mathcal{S}}\)
2) Let's show that \(ax' + by\) is a limit point. We need to show that for any ball \(B_\epsilon(ax' + by)\) there is some \(s \in S\) such that \(s \in B_\epsilon(ax' + by)\). In other words we must show that there is some \(s\) such that \(|(ax' + by) - s| < \epsilon\). Since \(x'\) is a limit point we can find some \(x\) such that \(|x' - x| < \frac{\epsilon}{a}\). Therefore take such an \(x\) and let \(s = ax + by\). Then \(|(ax' + by) - (ax + by)| = |ax' - ax| < \epsilon\). Therefore \(ax' + by\) is indeed a limit point of \(S\) and therefore \(ax' + by \in \overline{\mathcal{S}}\)

Answer 4

Does that make sense so far?

Answer 5

For 3 we can do something similar approximating a limit point ax' + by' via some point ax + by in S.

Answer 6

\(\bar S\) would be the smallest closed set that contains S , right ?
we need to show that \(\bar S=S\) , that would give \(\bar S\) is also open ,
2 direction proof :-
a) \(S \subset \bar S\) done
b) \(\bar S \subset S\) "we need to show that S is closed , then done

Answer 7

That doesn't make sense. Why would it be true that \(S =\overline{S}\). This is not true in general...

Answer 8

in general its not true , but ur qustion is asking about if S closure is a subspace which means if S closure is an open in ur space , so consider the properties

Answer 9

I'll give you an example to make this concrete. Let \(\mathcal{X}\) be the set of simple functions on \([0, 1]\) then the closure is the set of \(L^1\) functions on \([0, 1]\)

Answer 10

The set of simple functions is certainly not the same as the set of \(L^1\) functions but the simple functions are dense in the set of \(L^1\) functions.

Answer 11

The set of simple functions is a vector subspace because you can add simple functions and get a simple function.

Answer 12

ohkk , i dint notice that >.<

Answer 13

@SithsAndGiggles

Answer 14

I don't know near enough about the topic to help, but you can find a nice discussion of the proof here:
http://math.stackexchange.com/questions/202369/is-closure-of-linear-subspace-of-x-is-again-a-linear-subspace-of-x

Answer 15

It looks like I was on the right track. The third case looks like it requires a bit more work than the second though.