Lattice and Boolean Algebra over vector space (2) ----- inclusion-exclusion theorem and measure space

Math Essays: Lattice and Boolean Algebra over vector space (1) (wuyulanliulongblog.blogspot.com)

Consider a finite dimension inner product space $V$$V$, denote the lattice as $L_V$$L_V$.

The basis set is given by $S:=\{v_1,v_2,v_3,...,v_{n-1},v_n\}$$S:=\{v_1,v_2,v_3,...,v_{n-1},v_n\}$.

We know that

$\mathrm{Span}(A\cup B)=\mathrm{Span}(A)+\mathrm{Span}(B)$$\mathrm{Span}(A\cup B)=\mathrm{Span}(A)+\mathrm{Span}(B)$

$\mathrm{Span}(\mathrm{A}\cap \mathrm{B})=\mathrm{Span}(A)\cap \mathrm{Span}(B)$$\mathrm{Span(A\cap B)}=\mathrm{Span}(A)\cap\mathrm{Span}(B)$

Gives a lattice homomorphism

We know that we can have inclusion-exclusion theorem on $P(S)$$P(S)$

Math Essays: The connection between Principle of Inclusion-Exclusion and Combination (wuyulanliulongblog.blogspot.com)

Math Essays: Measure space and inclusion-exclusion Theorem (wuyulanliulongblog.blogspot.com)

$\begin{array}{c}\left|\bigcup _{i=1}^n{A}_i\right|=\sum _{i=1}^n|A_i|-\sum _{1\le i<j\le n}|A_i\cap A_j|+....+(-1{)}^{n+1}|A_1\cap A_2\cap ...\cap A_{n-1}\cap A_n|\end{array}$$\left|\bigcup_{i=1}^n A_i\right|=\sum_{i=1}^n|A_i|-\sum_{1\le i<j\le n}|A_i\cap A_j|+....+(-1)^{n+1}|A_1\cap A_2\cap...\cap A_{n-1}\cap A_n|\\$

Since we $\mathrm{\forall }A\in P(S),|A|=\dim \mathrm{Span}(\mathrm{A})$$\forall A\in P(S),|A|=\dim\mathrm{Span(A)}$

Thus we can have inclusion-exclusion theorem for $\dim$$\dim$as well

$\dim \sum _{i=1}^n\mathrm{Span}(A_i)$$\dim\sum_{i=1}^n\mathrm{Span}(A_i)$

$\begin{array}{c}=\sum _{i=1}^n\dim \mathrm{Span}(A_i)-\sum _{1\le i<j\le n}\dim \mathrm{Span}(A_i)\cap \mathrm{Span}{A}_j+...+(-1{)}^{n+1}\dim \bigcap _{i=1}^n\mathrm{Span}(A_i)\end{array}$$=\sum_{i=1}^n\dim\mathrm{Span}(A_i)-\sum_{1\le i<j\le n}\dim\mathrm{Span}(A_i)\cap\mathrm{Span}A_j+...+(-1)^{n+1}\dim\bigcap_{i=1}^n\mathrm{Span}(A_i)\\$

For example, in previous essay, we deduce this $\dim (H+K)=\dim (H)+\dim (K)-\dim (H\cap K)$$\dim(H+K)=\dim(H)+\dim(K)-\dim(H\cap K)$

from the third isomorphism theorem (it is the third isomorphism theorem, not the second, my bad, that is a typo)

But it can be viewed as a kind of inclusion-exclusion theorem as well in some condition!

From this we can see something more general and interesting.

Consider a finite dimension inner product space $V$$V$,

Define the $\sigma -\mathrm{A}\mathrm{l}\mathrm{g}\mathrm{e}\mathrm{b}\mathrm{r}\mathrm{a}$$\sigma-\rm Algebra$ as the $P(S)$$P(S)$, $S$$S$ is basis set of $V$$V$ and the measure function $\mu (A):=\dim \mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}(\mathrm{A})$$\mu(A):=\dim\rm Span(A)$

$P(S)$$P(S)$ is a sigma algebra is obviously, and $\dim \circ \mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}$$\dim\circ\rm Span$ is a measure function since

$†.\dim \circ \mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}(\mathrm{\varnothing })=\dim (0)=0$$\dagger.\dim\circ\rm Span(\empty)=\dim(0)=0$

$†.$$\dagger.$ $\begin{array}{c}\dim \circ \mathrm{Span}\left(\underset{i=1}{\overset{n}{⨆}}{A}_i\right)=\sum _{i=1}^n\dim \mathrm{Span}(A_i)\end{array}$$\dim\circ\mathrm{Span}\left(\bigsqcup_{i=1}^{n}A_i\right)=\sum_{i=1}^n\dim\mathrm{Span}(A_i)\\$

Consider the measure space $(S,P(S),\dim \circ \mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n})$$(S,P(S),\dim\circ\rm Span)$

And see the detail of this essay, using measure and integration to prove the inclusion-exclusion theorem

Math Essays: Measure space and inclusion-exclusion Theorem (wuyulanliulongblog.blogspot.com)