2
$\begingroup$

Show that every subspace of $\mathbb{R^n}$ is closed.

I'm not sure how to do this or even what closed means. I don't even have a starting point. Any hints or solutions are greatly appreciated.

$\endgroup$

closed as off-topic by Jonas Meyer, user99914, Daniel, Claude Leibovici, Macavity Jun 14 '15 at 5:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jonas Meyer, Community, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ I assume you mean vector subspace? If it's only a topological subspace then it's extremely false. $\endgroup$ – Matt Samuel Jun 4 '15 at 19:03
  • $\begingroup$ yeah, vector subspace $\endgroup$ – 1233dfv Jun 4 '15 at 19:04
2
$\begingroup$

A closed set contains all of its limit points.

$\endgroup$
2
$\begingroup$

A set is closed if and only if its complement is open. In $\mathbb{R}^n$ (or any metric space), a set $U$ is open if and only if for any $x\in U$ there exists an $\epsilon>0$ such that all points of distance less than $\epsilon$ from $x$ are contained in $U$.

$\endgroup$
2
$\begingroup$

Let $S$ be the subspace. Let $v_1,...,v_p$ be a basis for the subspace, and let $v_{p+1},...,v_n$ be such that $v_1,...,v_n$ form a basis for $\mathbb{R}^n$.

Define the linear map $\phi:\mathbb{R}^n \to \mathbb{R}^{n}$ by $\phi(v_k) = \begin{cases} 0, & k \le p \\ v_k, & k > p\end{cases}$, and note that $S = \ker \phi$. Since $\phi$ is continuous, it follows that $S$ is closed since $\ker \phi = \phi^{-1} \{0\}$ and $\{0\}$ is closed.

$\endgroup$
1
$\begingroup$

If $M$ is a $k$-dimensional subspace of $\mathbb{R}^{n}$, then you can use Gram-Schmidt to find an orthonormal basis $\{e_1,e_2,\cdots,e_k\}$ of $M$, complete this to a full orthonormal basis $\{e_1,e_2,\cdots,e_k,e_{k+1},\cdots,e_{n}\}$ and create an isometric isomorphism $$ U : \mathbb{R}^{n}\rightarrow \mathbb{R}^{n} $$ that maps $(x_1,x_2,\cdots,x_k,0,0,\cdots,0)$ onto $M$. So the problem reduces to showing that this second description, which is really $\mathbb{R}^{k}$, is closed.

$\endgroup$
0
$\begingroup$

Let $S$ be a linear subspace of $\mathbb{R}^n$. Consider a sequence $\{x_n\}_{n\in\mathbb{N}}$ of points in $S$ which converges to a point $y\in\mathbb{R}^n$. Show that in fact $y$ must lie in $S$. (Here I am using a characterization of "closed" which is equivalent to that described in the other answers).

You could use the continuity of any projection map to the subspace, if that's available to you or easy to prove. You could also show it directly by just doing the analysis. (Hint: the finite-dimensionality should matter)

$\endgroup$
0
$\begingroup$

I think the other answers are a bit high brow. You don't even know what a closed or open subset is.

The question seems to be asking, show that every vector subspace is topologically closed.

In $\mathbb{R}^n$ we may think of topologically closed sets as those that contain their boundary elements. This is a specific case of the definition given earlier talking about containing 'limit points'. An open subset is one that doesn't contain any of it's boundary elements.

An example in $\mathbb{R}$ is the interval $[0,1]$ is closed because it contains $0$ and $1$. These are the boundary elements because any open subset in $\mathbb{R}$ containing 0 must contain points outside of $[0,1]$ and the same is true of $1$.

The set $(0,1)$ containing all points between $0,1$ but not $0,1$ is open because it doesn't contain any of it's boundary points.

In higher dimensions, we can think of an open 'interval' as any 'blob' sitting inside of all of the dimensions of the space. So in $mathbb{R}^2$ this would be any section of the plane enclosed by a line. But not a line or a dot or any combination of lines or dots. An example might be a square on the plane without boundary.

We want to show that a vector subspace is a closed set. So that is, it contains it's boundary elements.

In $\mathbb{R}^n$ a set of boundary elements will itself be a closed set, because any open subset containing elements of this will contain elements of the boundary and elements outside the boundary. Therefore a boundary set is it's own boundary set, and contains itself and so is closed. And we'll show that a vector subspace is it's own boundary set.

Suppose we have $k$ linearly independent vectors $x_1,...,x_k\in \mathbb{R}^n$. When we talk the span of these vectors we get a vector subspace. $$S = span(x_1,...,x_k)$$ All vector subspaces are formed with such a span, and this isn't unique.

We take a point in $x \in S$, and by the definition of $span(x_1,...,x_k)$, $x = a_1x_1+...+a_kx_k$, with $a_1,..,a_k \in \mathbb{R}$.

We consider an arbitrary open subset $U_x$ on the point $x$. This contains elements of our subspace obviously.

I'll prove the above statement. We consider a ball centred at $x$ $B_x \subset U_x$ and take the radius of the ball $r$. We take $\epsilon = \frac{r}{2}$ and then consider the point $x+\epsilon x_1 = (a_1+\epsilon)x_1+...+a_kx_k$. This is still in $S$ by definition of a span, and in $U_x$ because it is in $B_x$ because the distance between $x$ and $x+\epsilon x_1$ is less than the radius of $B_x$.

We also want to show that for all $U_x$ there is a point not in our subspace. This is easy. Since $S$ is a strict subspace, we can take $x_{k+1}$ linearly independent from $x_1,...,x_k$. By the same construction we can consider a ball $B_x \subset U_x$ and $x+\epsilon x_{k+1} \not \in S$.

Thus for all $x \in S$ and all $U_x$ such that $x \in U_x$, $x$ is a boundary point of $S$. Therefore $S$ contains it's boundary, and $S$ is closed, for all subspaces of $\mathbb{R}^n$ formed by taking the span of vectors. Which is all vector subspaces.

So all vector subspaces are topologically closed.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.