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 Limit Theory of set sequence
04-10-2017, 06:09 PM
Post: #1
 elim Moderator Posts: 580 Joined: Feb 2010 Reputation: 0
Limit Theory of set sequence
Definition $A\Delta B:=(A-B)\cup(B-A)$ is called
$\quad$the symmetric difference of $A,B$.

Remark Binary Operator $\Delta$(Symmetric Difference)
$\quad$has the following algebraic properties:
$\;\;\;1)\;\;A\Delta\varnothing = A,$
$\;\;\;2)\;\;A\Delta A = \varnothing,$
$\;\;\;3)\;\;A^c\Delta B^c = A\Delta B,$
$\;\;\;4)\;\;A\Delta B = B\Delta A,$
$\;\;\;5)\;\;(A\Delta B)\Delta C = A\Delta(B\Delta C).$

Proof $1)\sim 4)$ are trivial. So we turn directly to 5):
$\quad(A\Delta B)\Delta C \underset{\,}{=}$
$\qquad =(((A-B)\cup(B-A))-C)\cup(C-((A-B)\cup(B-A)))$
$\qquad =((A\cap B^c\cap C^c)\cup(A^c\cap B\cap C^c))\cup(C\cap((A^c\cup B)\cap(A\cup B^c)))$
$\qquad \underset{\,}{=}((A\cap B^c\cap C^c)\cup(A^c\cap B\cap C^c))\cup(C\cap((A^c\cap B^c)\cup(A\cap B)))$
$\qquad \underset{\,}{=}(A\cap B^c\cap C^c)\cup(A^c\cap B\cap C^c)\cup(A^c\cap B^c\cap C)\cup(A\cap B\cap C)$
Now$\underset{\,}{\;}A\Delta(B\Delta C)=(B\Delta C)\Delta A$ Plug the latter to the above we get
$\qquad \underset{\,}{=}(B\cap C^c\cap A^c)\cup(B^c\cap C\cap A^c)\cup(B^c\cap C^c\cap A)\cup(B\cap C\cap A)$
$\qquad \underset{\,}{=}(A\cap B^c\cap C^c)\cup(A^c\cap B\cap C^c)\cup(A^c\cap B^c\cap C)\cup(A\cap B\cap C)$
$\qquad = (A\Delta B)\Delta C.\quad\square$

Definition Let $A,\,\{A_n\}$be set and a sequenct of sets.
$\quad$Say $\{A_n\}$ converges to $A$, write $\displaystyle{\lim_{n\to\infty}A_n = A}$, if
$(1)\qquad\qquad\forall x\,\exists N\,\forall n{\small >}N\;(x\not\in A_n\Delta A).$
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