Is it possible that a subspace is not a vector space

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tze liu
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<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?
 
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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?
 
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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}##
 
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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
 
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.
 
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
 
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)?
 
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.
 
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
 
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.
 
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