# Is Every Non-Empty Set Closed Under Addition a Subspace in F2?

• ilyas.h
In summary: W}therefore, 0_{W} is the zero element and is contained in W (since W is non-empty).Proof of additive inverses:We know that 0x = 0_{W} from the previous proof. If we take the additive inverse of 0x we get:(0x)^{-1} = (0_{W})^{-1}since (0x)^{-1} exists (because x is in the subspace W and W is a subspace of V) then (0_{W})^{-1} must also exist. Therefore, every element in W has an
ilyas.h

## Homework Statement

Let $F_{2} = {0, 1}$ denote a field with 2 elements.

Let V be a vector space over $F_{2}$. Show that every non-empty set W of V which is closed under addition is a subspace of V.

## The Attempt at a Solution

subspace axioms: 0 elements, closed under scalar multiplication, closed under vector addition.

We can skip the latter axiom as it's given in the question.

proof of 0 element:

0, 1 ∈ F_2
x ∈ W

$0x = 0_{W}$

therefore there exists a zero element.

proof of scalar multiplication:

0, 1 ∈ F_2
x ∈ W

$1x = x$

this is true due to the scalar multiplication identity.

----------------------------------

I believe this could be wrong, I feel as though I am missing something. Thanks.

Last edited:
ilyas.h said:

## Homework Statement

Let $F_{2} = {0, 1}$ denote a field with 2 elements.

Let V be a vector space over $F_{2}$. Show that every non-empty set W of V which is closed under addition is a subspace of V.

## The Attempt at a Solution

subspace axioms: 0 elements, closed under scalar multiplication, closed under vector addition.

We can skip the latter axiom as it's given in the question.

proof of 0 element:

0, 1 ∈ F_2
x ∈ W

$0x = 0_{W}$

therefore there exists a zero element.

proof of scalar multiplication:

0, 1 ∈ F_2
x ∈ W

$1x = x$

this is true due to the scalar multiplication identity.

----------------------------------

I believe this could be wrong, I feel as though I am missing something. Thanks.

The words describing what you are doing could be clearer, but basically all the steps are there. All you need to do is prove that W contains the zero vector (which you did) and is closed under scalar multiplication. You showed that if x is in W then 1x is in W. For completeness you might want to mention why 0x is in W. It really is a pretty easy problem.

Actually thinking about it again, you also need to show that there are additive inverses. 'Closed under addition' doesn't imply that.

Dick said:
The words describing what you are doing could be clearer, but basically all the steps are there. All you need to do is prove that W contains the zero vector (which you did) and is closed under scalar multiplication. You showed that if x is in W then 1x is in W. For completeness you might want to mention why 0x is in W. It really is a pretty easy problem.

Actually thinking about it again, you also need to show that there are additive inverses. 'Closed under addition' doesn't imply that.

I think I've found a more robust proof for scalar multiplication:

0, 1 in F_2
x in W

consider all possible combinations:

0x, 0x + 1x, 1x + 1x, 0x + 0x, 1x + 1x + 1x

$0x = 0_{W}$
$0x + 1x = 0_{W} + x = x$
$1x + 1x = (1 + 1)x = 0x = 0_{W}$
$0x + 0x = 0_{W}$
$1x + 1x + 1x = (1 + 1)x + 1x = 0x + 1x = 0_{W} + x = x$
all of them belong to W. But if you compare the second and last line you can see they are the same, and so we're going around in circles, I've shown all possible outcomes between the (very small) Field and the subspace W.

I don't think I need to show that it contains the zero element since it's non-empty and so by definition, it contains the zero element.

ilyas.h said:

## Homework Statement

Let $F_{2} = {0, 1}$ denote a field with 2 elements.

Let V be a vector space over $F_{2}$. Show that every non-empty set W of V which is closed under addition is a subspace of V.

## The Attempt at a Solution

subspace axioms: 0 elements, closed under scalar multiplication, closed under vector addition.

We can skip the latter axiom as it's given in the question.

proof of 0 element:

0, 1 ∈ F_2
x ∈ W

$0x = 0_{W}$

therefore there exists a zero element.

$W$ is a subset of $V$. If $W$ is to be a subspace of $V$ then you need $0_V \in W$.

You have to show why the condition that $W$ is closed under vector addition requires that $0_V \in W$, so that if $w \in W$ then both $1w = w \in W$ and $0w = 0_V \in W$.

Let $w \in W$. Since $W$ is closed under addition, we have $w + w \in W$. What do you know about $w + w = (1 + 1)w$?

ilyas.h said:
I think I've found a more robust proof for scalar multiplication:

0, 1 in F_2
x in W

consider all possible combinations:

0x, 0x + 1x, 1x + 1x, 0x + 0x, 1x + 1x + 1x

$0x = 0_{W}$
$0x + 1x = 0_{W} + x = x$
$1x + 1x = (1 + 1)x = 0x = 0_{W}$
$0x + 0x = 0_{W}$
$1x + 1x + 1x = (1 + 1)x + 1x = 0x + 1x = 0_{W} + x = x$
all of them belong to W. But if you compare the second and last line you can see they are the same, and so we're going around in circles, I've shown all possible outcomes between the (very small) Field and the subspace W.

I don't think I need to show that it contains the zero element since it's non-empty and so by definition, it contains the zero element.

You've added a bunch of unnecessary stuff and removed some important stuff. It's not adding robustness, it's adding confusion. Could you say what you want to prove (and why!) before the actual proof. "proof for scalar multiplication" doesn't do it. WHAT about scalar multiplication?

Dick said:
You've added a bunch of unnecessary stuff and removed some important stuff. It's not adding robustness, it's adding confusion. Could you say what you want to prove (and why!) before the actual proof. "proof for scalar multiplication" doesn't do it. WHAT about scalar multiplication?

show that W is a subspace of the vector space V (and V is a vector space over the Field F = {0, 1}).

1.x = x

since 1 is in the field and x is in W (and W is a subspace of V) then we can use the scalar multiplication identity from V. Ergo, scalar multiplication holds for this subspace.

Proof of zero element:

1x + 1x = (1+1)x (distributivity law from the vector space, V, onto subspace, W).#

(1+1)x = 0x = 0_w

since 1x + 1x ∈ W, since elements in a subspace added together are also in the subspace, then 0_w must also be in the subspace. Therefore 0 element exists.
What do you think of my proof now? I am trying to connect all the dots. Thanks.

Last edited:
ilyas.h said:
show that W is a subspace of the vector space V (and V is a vector space over the Field F = {0, 1}).

1.x = x

since 1 is in the field and x is in W (and W is a subspace of V) then we can use the scalar multiplication identity from V. Ergo, scalar multiplication holds for this subspace.

Proof of zero element:

1x + 1x = (1+1)x (distributivity law from the vector space, V, onto subspace, W).#

(1+1)x = 0x

since 1x + 1x ∈ W, since elements in a subspace added together are also in the subspace, then 0_w must also be in the subspace. Therefore 0 element exists.
What do you think of my proof now? I am trying to connect all the dots. Thanks.

Getting better! The first part could be clearer. What you need to prove there is that the set W is closed under scalar multiplication. I.e. if x is in W then 0x and 1x are also in W. You might want to prove that the zero vector is in W first. I'd also suggest you comment on why additive inverses of elements of W are also in W.

## 1. What are field axioms?

Field axioms are a set of mathematical rules that define the properties of a field, which is a set of numbers that can be added, subtracted, multiplied, and divided.

## 2. How are field axioms used in mathematics?

Field axioms serve as the foundation for algebraic structures such as vector spaces, which are used to model mathematical concepts in fields such as physics, engineering, and economics.

## 3. Can you give an example of a field axiom?

One example of a field axiom is the commutative property of multiplication, which states that the order of the factors does not affect the result. For example, in the real numbers, a x b = b x a.

## 4. What is a subspace?

A subspace is a subset of a vector space that satisfies all of the vector space axioms, making it a vector space in its own right. It must contain the zero vector, be closed under addition and scalar multiplication, and contain all linear combinations of its elements.

## 5. How are subspaces related to field axioms?

Subspaces rely on the field axioms because they are defined using vectors, which are elements of a vector space. Without the field axioms, the properties of vector spaces and subspaces would not hold, making them invalid mathematical structures.

• Calculus and Beyond Homework Help
Replies
7
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
0
Views
543
• Calculus and Beyond Homework Help
Replies
10
Views
2K
• Calculus and Beyond Homework Help
Replies
15
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
2K
• Calculus and Beyond Homework Help
Replies
8
Views
1K
• Calculus and Beyond Homework Help
Replies
7
Views
1K