1003: For Vectors Space and Set of Sub-'Vectors Spaces', Intersection of Set Is Sub-'Vectors Space'

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description/proof of that for vectors space and set of sub-'vectors spaces', intersection of set is sub-'vectors space'

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of %field name% vectors space.

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Target Context

• The reader will have a description and a proof of the proposition that for any vectors space and its any set of sub-'vectors space's, the intersection of the set is a sub-'vectors space'.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\text{ the fields }\}$
$V^{\prime }$: $\in \{\text{ the vectors spaces over }F\}$
$\{V_{\beta }|\beta \in B\}$: $B\in \{\text{ the uncountable index sets }\}$, $V_{\beta }\in \{\text{ the sub-'vectors space's of }{V}^{\prime }\}$
$V$: $=\cap \{V_{\beta }|\beta \in B\}$
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Statements:
$V\in \{\text{ the sub-'vectors space's of }{V}^{\prime }\}$
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2: Proof

Whole Strategy: Step 1: see that $V$ satisfies the requirements to be a vectors space.

Step 1:

1) for any elements, $v_1,v_2\in V$, $v_1+v_2\in V$ (closed-ness under addition): $v_j\in V_{\beta }$ for each $\beta$, so, $v_1+v_2\in V_{\beta }$ for each $\beta$, so, $v_1+v_2\in V$.

2) for any elements, $v_1,v_2\in V$, $v_1+v_2=v_2+v_1$ (commutativity of addition): $v_1+v_2=v_2+v_1$ on ambient $V^{\prime }$, so, $v_1+v_2=v_2+v_1$ on $V$.

3) for any elements, $v_1,v_2,v_3\in V$, $(v_1+v_2)+v_3=v_1+(v_2+v_3)$ (associativity of additions): $(v_1+v_2)+v_3=v_1+(v_2+v_3)$ on ambient $V^{\prime }$, so, $(v_1+v_2)+v_3=v_1+(v_2+v_3)$ on $V$.

4) there is a 0 element, $0\in V$, such that for any $v\in V$, $v+0=v$ (existence of 0 vector): $0\in V_{\beta }$ for each $\beta$, so, $0\in V$.

5) for any element, $v\in V$, there is an inverse element, $v^{\prime }\in V$, such that $v^{\prime }+v=0$ (existence of inverse vector): $v_{\beta }^{\prime }\in V_{\beta }$ for each $\beta$ (we do not rule out the possibility that $v_{\beta }^{\prime }$ depends on $\beta$), but $v_{\beta }^{\prime }$ is an inverse also on $V^{\prime }$ because $v_{\beta }^{\prime }+v=0$ there and the inverse is unique because from $v_{\beta }^{\prime }+v=0$, $v_{\beta }^{\prime }+v-v=0-v$, which implies that $v_{\beta }^{\prime }=-v:=v^{\prime }$, so, $v^{\prime }\in V_{\beta }$ for each $\beta$, so, $v^{\prime }\in V$ with $v^{\prime }+v=0$.

6) for any element, $v\in V$, and any scalar, $r\in F$, $r.v\in V$ (closed-ness under scalar multiplication): $v\in V_{\beta }$ for each $\beta$, so, $r.v\in V_{\beta }$ for each $\beta$, so, $r.v\in V$.

7) for any element, $v\in V$, and any scalars, $r_1,r_2\in F$, $(r_1+r_2).v=r_1.v+r_2.v$ (scalar multiplication distributability for scalars addition): $(r_1+r_2).v=r_1.v+r_2.v$ on $V$, because it is so on ambient $V^{\prime }$.

8) for any elements, $v_1,v_2\in V$, and any scalar, $r\in F$, $r.(v_1+v_2)=r.v_1+r.v_2$ (scalar multiplication distributability for vectors addition): $r.(v_1+v_2)=r.v_1+r.v_2$ on $V$, because it is so on ambient $V^{\prime }$.

9) for any element, $v\in V$, and any scalars, $r_1,r_2\in F$, $(r_1{r}_2).v=r_1.(r_2.v)$ (associativity of scalar multiplications): $(r_1{r}_2).v=r_1.(r_2.v)$ on $V$, because it is so on ambient $V^{\prime }$.

10) for any element, $v\in V$, $1.v=v$ (identity of 1 multiplication): $1.v=v$ on $V$, because it is so on ambient $V^{\prime }$.

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References

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