Limit Theory of set sequence - Printable Version +- OMath! (http://math.elinkage.net) +-- Forum: Math Forums (/forumdisplay.php?fid=4) +--- Forum: Math Foundations (/forumdisplay.php?fid=13) +--- Thread: Limit Theory of set sequence (/showthread.php?tid=728) Limit Theory of set sequence - elim - 04-10-2017 06:09 PM 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 $(1)\qquad\qquad\forall x\,\exists N\,\forall n{\small >}N\;(x\not\in A_n\Delta A).$