description/proof of that for vectors space and set of sub-'vectors spaces', intersection of set is sub-'vectors space'
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of %field name% vectors space.
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'.
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.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\text{ the fields }\}\)
\(V'\): \(\in \{\text{ the vectors spaces over } F\}\)
\(\{V_\beta \vert \beta \in B\}\): \(B \in \{\text{ the uncountable index sets }\}\), \(V_\beta \in \{\text{ the sub-'vectors space's of } V'\}\)
\(V\): \(= \cap \{V_\beta \vert \beta \in B\}\)
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Statements:
\(V \in \{\text{ the sub-'vectors space's of } V'\}\)
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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 \(V'\), 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 \(V'\), 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, \(v' \in V\), such that \(v' + v = 0\) (existence of inverse vector): \(v'_\beta \in V_\beta\) for each \(\beta\) (we do not rule out the possibility that \(v'_\beta\) depends on \(\beta\)), but \(v'_\beta\) is an inverse also on \(V'\) because \(v'_\beta + v = 0\) there and the inverse is unique because from \(v'_\beta + v = 0\), \(v'_\beta + v - v = 0 - v\), which implies that \(v'_\beta = - v := v'\), so, \(v' \in V_\beta\) for each \(\beta\), so, \(v' \in V\) with \(v' + v = 0\).
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 \(V'\).
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 \(V'\).
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 \(V'\).
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 \(V'\).
