initial topology with respect to norm

Mar 2014
6
1
Austria
Hello,

Suppose \(\displaystyle (X,\|.\|)\) is a normed vector space. Let \(\displaystyle (X,\mathcal{T})\) be the initial topology with respect to \(\displaystyle \|.\|: X \rightarrow [0,\infty)\). Then the topology induced by the norm is not equal to \(\displaystyle \mathcal{T}\).

I've been thinking about this for some time now but I'm having real trouble proving it. Any help would be appreciated.

Best Regards,
Stiwan
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
Hey Stiwan.

This is probably bad intuition, but I'd intuitively conclude that both are isomorphic topology wise. I'm just picture this geometrically since basically the norm is continuous and has a specific topology (this is for any norm, not a particular one if it follows the axioms).

Maybe you should look at specific topological properties of the norm and see if you can induce some kind of appropriate isomorphism (or show that you can't).
 
Mar 2014
6
1
Austria
What do you mean by "isomorphic topology wise"? That they are homeomorphic? "The norm is continuous"... which topologies are you referring to? This does not hold in general, e.g. if you define it as a map from \(\displaystyle (X,\{\emptyset, X\})\) to\(\displaystyle ([0,\infty),|.|)\). I think the right way to tackle this problem would be to find a topology \(\displaystyle \mathcal{T'}\) such that the norm is continuous so that there is a set that is open in \(\displaystyle (X,\|.\|)\) but is not open in \(\displaystyle (X,\mathcal{T'})\). I had some ideas (like defining a metric \(\displaystyle d(x,y) := \|x\| + \|y\|\) if \(\displaystyle x\neq y\) and \(\displaystyle d(x,y) := 0\) if \(\displaystyle x = y\)), but none of them really worked.
 

Deveno

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Mar 2011
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Tejas
Consider the topology:

$\tau = \langle\{B_r(0):r \in [0,\infty)\} \cup \{A_{r,s}(0):r< s \in [0,\infty)\}\rangle$ (these sets form a BASE).

where:

$B_r(0) = \{x \in X: \|x\| < r\}$ and:

$A_{r,s}(0) = \{x \in X: r < \|x\| < s\}$

Show that this topology makes the norm continuous, but is not consistent with vector addition: in other words, $B_r(x)$ is in the topology induced by the norm, but not in $\tau$

(here, $B_r(x) = \{y \in X: \|x - y\| < r\}$).

Intuitively, all we need to make the norm continuous is "neighborhoods centered at 0", but to make the norm respect the vector space addition, we also need "translates of these neighborhoods".
 
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Mar 2014
6
1
Austria
Consider the topology:


Show that this topology makes the norm continuous, but is not consistent with vector addition: in other words, $B_r(x)$ is in the topology induced by the norm, but not in $\tau$

(here, $B_r(x) = \{y \in X: \|x - y\| < r\}$).

Intuitively, all we need to make the norm continuous is "neighborhoods centered at 0", but to make the norm respect the vector space addition, we also need "translates of these neighborhoods".
I don't really see why all we need to make the norm continuous is "neighborhoods centered at 0". How would you prove that the norm is continuous in this case? Let's look at a special case to make things easier:

Suppose \(\displaystyle (X,\|.\|) = (\mathbb{R},|.|)\). Then, the preimage of \(\displaystyle B_r(x) := \{y \in R : |x-y|< r\}\) is equal to \(\displaystyle B_r(x) \cup B_r(-x)\). How would you show that any set of this type is equal to a union of sets that are elements of your base?

Anyway, I think I've found quite an easy way to prove it: In general, it holds that the preimage of any subset of \(\displaystyle [0,\infty)\) with respect to \(\displaystyle \|.\|\) is symmetric. Because the preimages of open subsets of \(\displaystyle [0,\infty)\) form a subbase of the initial topology and the union and intersection of symmetric sets is always a symmetric set itself, any set that is open in the initial topology must be symmetric. As $B_r(x) = \{y \in X: \|x - y\| < r\}$ is not symmetric if $ x \neq 0$ but open in the topology induced by the norm, the topologies can not be equal to each other.
 

Deveno

MHF Hall of Honor
Mar 2011
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I don't really see why all we need to make the norm continuous is "neighborhoods centered at 0". How would you prove that the norm is continuous in this case? Let's look at a special case to make things easier:

Suppose \(\displaystyle (X,\|.\|) = (\mathbb{R},|.|)\). Then, the preimage of \(\displaystyle B_r(x) := \{y \in R : |x-y|< r\}\) is equal to \(\displaystyle B_r(x) \cup B_r(-x)\). How would you show that any set of this type is equal to a union of sets that are elements of your base?

Anyway, I think I've found quite an easy way to prove it: In general, it holds that the preimage of any subset of \(\displaystyle [0,\infty)\) with respect to \(\displaystyle \|.\|\) is symmetric. Because the preimages of open subsets of \(\displaystyle [0,\infty)\) form a subbase of the initial topology and the union and intersection of symmetric sets is always a symmetric set itself, any set that is open in the initial topology must be symmetric. As $B_r(x) = \{y \in X: \|x - y\| < r\}$ is not symmetric if $ x \neq 0$ but open in the topology induced by the norm, the topologies can not be equal to each other.
The sets I gave for $\tau$ ARE symmetric (usually, see below). Note that the pre-image you gave is the same as $A_{|x|-r,|x|+r}(0)$ (the "A" in my sets is supposed to suggest "annulus" in the same vein as "B" is supposed to suggest "ball").

I avoided the term "symmetric" because in the case that $X = [0,\infty)$ itself with the usual norm, the pre-image of $\|\cdot\|$ (which is now the identity function) of any set is now itself, whereas the definition of $B_r(0),A_{r,s}(0)$ remains unchanged. This can be made into a vector space by taking the "vector addition"

$x \oplus y = xy$

and "scalar multiplication":

$c \odot x = e^cx$

In this "weird" vector space, the real number 1 is now the 0-vector, and the pre-images of an open set in $X$ aren't "symmetric" in the usual sense.