Is it possible that a subspace is not a vector space

Click For Summary

Homework Help Overview

The discussion revolves around the concept of subspaces in vector spaces, specifically whether a subspace can exist that does not fulfill all the axioms required for a vector space. Participants explore the definitions and properties of subspaces, including closure under addition and scalar multiplication.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants question the necessity of all vector space axioms for subspaces and discuss examples of subsets that are closed under addition but do not qualify as subspaces. There is inquiry into the implications of closure properties and the relationship between subspaces and their parent vector spaces.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants provide clarifications on definitions, while others express concerns about potential confusion arising from earlier statements. There is no explicit consensus on the definitions or examples being debated.

Contextual Notes

Some participants reference specific subsets, such as the integers and the trivial subspaces, to illustrate their points. There is an emphasis on standard terminology and definitions within the context of vector spaces and subspaces.

tze liu
Messages
57
Reaction score
0
<Mentor's note: moved from general mathematics to homework. Thus no template.>

Prove subspace is only a subset of vector space but not a vector space itself.
Even a subspace follows closed under addition or closed under multiplication,however it is not necessary to follow other 8 axioms in vector space.
if some subspace follow closed under addition or closed under multiplication but don't follow Distributive axioms of c(x+y) = cx + cy(mean they cannot be added in this way),then it is a subspace but not vector space.
thank
right?
 
Physics news on Phys.org
You said for a subspace ##U## with ##x,y \in U## and ##c \in \mathbb{F}## you have ##c\cdot x\, , \,x+y \in U##.
With this, can you say something about ##cx+cy##? And why is ##c(x+y) \neq cx +cy## impossible?
 
If the subspace is a subspace of a vector space, then it has to inherit the other axioms from the parent space.
Otherwise: subspace of what?
 
  • Like
Likes   Reactions: tze liu
tze liu said:
Prove subspace is only a subset of vector space but not a vector space itself.
then what is a subspace by definition?
tze liu said:
Even a subspace follows closed under addition or closed under multiplication,however it is not necessary to follow other 8 axioms in vector space.
I can imagine such a set but it is not called subspace in accordance to standard terminology.
I mean that even in ##\mathbb{R}## there are subsets those are closed under + but they are not subspaces of the one dimensional vector space ##\mathbb{R}##
 
Last edited:
  • Like
Likes   Reactions: tze liu
zwierz said:
I mean that even in ##\mathbb{R}## there are subsets those are closed under + but they are not subspaces of the one dimensional vector space ##\mathbb{R}##
Well, there are the two trivial ones that are subspaces.
 
##\mathbb{Z}## is not a subspace
 
Who is talking about Z?
{0} and R are.
 
fresh_42 said:
You said for a subspace ##U## with ##x,y \in U## and ##c \in \mathbb{F}## you have ##c\cdot x\, , \,x+y \in U##.
With this, can you say something about ##cx+cy##? And why is ##c(x+y) \neq cx +cy## impossible?
what is the reason that closed under addition and closed under multiplication lead to the conclusion that c(x+y) = cx + cy?

x+y=y+x closed under addition
c(x+y)=c(y+x) closed under mutiplication
cx+cy is closed under add / muti
but these 3 properties does not give the conclusion that c(x+y)=cx+cy?
 
zwierz said:
then what is a subspace by definition?

I can imagine such a set but it is not called subspace in accordance to standard terminology.
I mean that even in ##\mathbb{R}## there are subsets those are closed under + but they are not subspaces of the one dimensional vector space ##\mathbb{R}##
subspace is a subset of vector space that follows the closed add/muti rule.
 
  • #10
mfb said:
Who is talking about Z?
it is just an example of a subset that is closed under + but it is not a subspace of the vector space ##\mathbb{R}## (over the field ##\mathbb{R}##)
tze liu said:
subspace is a subset of vector space that follows the closed add/muti rule.
seems that you use not standard definition
 
  • #11
tze liu said:
subspace is a subset of vector space that follows the closed add/muti rule.
Correct.
tze liu said:
but these 3 properties does not give the conclusion that c(x+y)=cx+cy?
How could they be different (in the subspace) and simultaneously equal (in the vector space)?
 
  • #12
tze liu said:
what is the reason that closed under addition and closed under multiplication lead to the conclusion that c(x+y) = cx + cy?
You inherit the definition of multiplication and addition from the parent space.@zwierz: Your earlier post, taken literally, claimed that no set closed under addition is a subspace of R. That is wrong.
There are sets closed under addition that are not subspaces, sure, but that is a weaker statement than you made.
 
  • #13
mfb said:
Your earlier post, taken literally, claimed that no set closed under addition is a subspace of R. That is wrong.
which post do you mean? cite it please
 
  • #14
This is a homework thread and the discussion drifts away from any help to the OP. Instead it is very likely starting to confuse him as wrong statements already have been made. The only purpose here is to help @tze liu and not to argue about vector spaces and its subspaces.
 
  • #15
fresh_42 said:
Instead it is very likely starting to confuse him as wrong statements already have been made
which wrong statements do you mean cite them please
 

Similar threads

Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 19 ·
Replies
19
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
2
Views
2K