TOPOLOGY SHORT NOTES 2

Neighborhood(nbd)
$A\; subset\; A\subseteq X\;is\;nbd\; of\;point\;x\in X,\;if\;\exists\; an$
$open\; set\;U\in \tau\;such\;that\;x\in U\subseteq A\;denoted\;by\;N_{x}$
$There\; may\; be\; several\; nbds\; of\; x\in X$
Neighborhood System
$The\;class\;of\;neighborhood\; of\; x\in X\; is\;denoted\; by\; N(x).$
Examples
$1.\; a\in \mathbb{R},\;then\;[a-\delta,a+\delta]\;is\;nbd\;of\; a.$
$2.\; In\; an\;Indiscrete\;topological\; space\;\tau=\{\phi,X\},\;only\;X\;is\;the\;nbd\;of\;all\;of\;its\;points$
$\maltese\; A \;subset A\subseteq X \;is\;open\;if\;and\;only\;if\;A\;is \;nbd\;of\;each\;of\;its\;point$
$\maltese\; N(x)\; of\;x\;in\;space\;X\;has\;properties$
                $\bullet \; If\; A\subseteq N(x)\;then\;x\in A$
                $\bullet \; If\; A,B\subseteq N(x)\;then\;A\cap B\in N(x)$
                $\bullet \; If\; A\subseteq N(x),A\subseteq B\;then\;B\in N(X)$
Limit/Cluster/Accumulation  Point
$A\;point\;x\in X\;is\;limit\;point\;of\;set\;A\subset X,\;iff\;every\;nbd\;of\;x,\; contains\;a\;point\;of\;A\;$
$other\;than\;x.i.e;$
$N_{x}\cap(A-\{x\})\neq \phi$
$or\; (N_{x} -\{x\})\cap A\neq\phi$
$A^{d}=\{x\in X:\forall N_{x},N_{x}\cap(A-\{x\})\}$
Examples
$1.\;A=\{0,1,1/2,1/3,...\} \;then\;A^{d}=\{0\}$
$2.\;B=\{1,1/2,1/3,...\} \;then\;B^{d}=\{0\}$
$3.\;C=(0,1)\;or\;(0,1],\;[0,1),\;[0,1]\;then\; C^{d}=[0,1]$
$4.\;E=\{0,1,2,3,...\}\; has\; no\; limit\; point$
Note
$\star If\;space\;X\;is\;discrete\;then\;any\;A\subseteq X\;has\;no\;limit\;point\;as\;it\;contains$
$\;singelton\;set\;of\;each\;x\in X.$
Closed Set
1. A is closed if its compliment is open
2. A is close iff derived set of A contain in A
NOTE:
● There are may be sets in X which are neither open nor close.
Examples:
1. In discrete space, every A contain in X is open as well as close
2. In Indiscrete space every A not equal to phi and A not equal to X is neither open nor close.
3. R is usual space
● Let A={1, 1/2, 1/3, …} is not closed as derived set of A does not contain in A
● Let B={0, 1/2, 1/3, …} closed as derived set of B contain in B
● Let C=(-infinity, 0)U(1, infinity) U(1, 1/2) U (1/2,1/3)U... is not closed as derived set of C does not contain in C