>> and some that do not. Interior and Boundary Points of a Set in a Metric Space Fold Unfold. $\quad S = \{ \bfx \in \R^n : |\bfx| = 2^{-j} \mbox{ for some }j\in {\mathbb N}\}$. endobj Interior, boundary, and closure Assume that $S\subset \R^n$ and that $\bfx$ is a point in $\R^n$. /ItalicAngle 0 More precisely, >> S^{int} \subset S \subset \bar S. 19 0 obj This can be done by choosing a point $\bfy$ of the form $\bfy = \bfa + t(\bfx - \bfa)$ and then adjusting $t$ suitably. /Resources << $\newcommand{\bfb}{\mathbf b}$ Show that $\cap_{j\ge 1} B(1+ 2^{-j}, {\bf 0}) = \{ \bfx\in \R^n : |\bfx| \le 1\}.$. Conversely, assume that $\partial S\subset S$. Or, equivalently, the closure of solid Scontains all points that are not in the exterior of S. A closed interval [a;b] ⊆R is a closed set since the set Rr[a;b] = (−∞;a)∪(b;+∞)is open in R. 5.3 Example. Finally, the statement that For example. Determine (without proof) whether the following sets are open, closed, neither, or both. If $A_1, A_2, \ldots$ is a sequence of subsets of $\R^n$, then The other topological structures like exterior and boundary have remain untouched. Prove that your answer is correct. /pgf@ca0.7 << \end{equation}. \end{equation}, There is some magnification beyond which, in your view-finder, endobj † The closure of A is the set c(A) := A[d(A).This set is sometimes denoted by A. >> If we want to prove these (not recommended, for the assertion about $\partial S$), we can do so as follows: $\newcommand{\ep}{\varepsilon}$. 15 0 obj B(r, \bfa) := \{ \bfx \in \R^n : |\bfx - \bfa|< r\}. In this section, we introduce the concepts of exterior and boundary in multiset topology. /pgfpat4 16 0 R This is the same as saying that /Parent 1 0 R the union of interior, exterior and boundary of a solid is the whole space. They are terms pertinent to the topology of two or $$ Deduce from problem 1 above and de Morgan's laws that if $A, B$ are closed subsets of $\R^n$ then $A\cup B$ and $A\cap B$ are closed. By applying the definitions, we can see that $B(\ep ,\bfx)\cap S\ne \emptyset$. /Length 53 >> /pgf@ca0.4 << $$. \mbox{ there exists }\ep>0\mbox{ such that }B(\ep, \bfx)\subset S^c. E��$^�.�DR����o�1�;�mV
��k����'72��x3[������W��b[Bs$4���Uo�0ڥ�|��~٠��u���-��G¸N����`_M�^ dh�;���XjR=}��F6sa��Lpd�,�)6��`cg�|�Kqc�R�����:Jln��(�6���5t�W;�2� �Z�F/�f�a�rpY��zU���b(�>���b��:;=TNH��#)o _ۈ}J)^?J�N��u��Ez��v|�UQz���AڡD�o���jaw.�:E�VB ���2��|����2[D2�� What is the interior of S? $\qquad \Box$, Theorem 2. $\newcommand{\bfu}{\mathbf u}$ Next, since $\partial S = \partial S^c$ and every point of $S^c$ belongs either to $(S^c)^{int}$ or $\partial(S^c)$, Assume that $\bfa\in \R^n$ and that $r>0$. /Length 1967 (S^c)^c = S. Next, consider an arbitrary point $\bfx$ of $S$. /Type /Page %PDF-1.3 \mbox{ there exists }\ep>0\mbox{ such that }B(\ep, \bfx)\subset S. �06l��}g �i���X%ײַ���(���H�6p�������d��y~������,y�W�b�����T�~2��>D�}�D��R����ɪ9�����}�Y]���`m-*͚e������E�!��.������u�7]�.�:�3�cX�6�ܹn�Tg8أ���:Y�R&�
� �+oo�o�YM�R���� /ca 0.25 This could mean questions completely unlike the ones below but at a similar level of difficulty. 8 0 obj Prove that if $A, B$ are open subsets of $\R^n$ then $A\cup B$ and $A\cap B$ are open. /MediaBox [ 0 0 612 792 ] << This will mostly be unnecessary, $\quad S = \{(x,y)\in \R^2 : x>0 \mbox{ and } y\ge 0\}$. $\bfx \in S^{int}$. Interior, closure, and boundary We wish to develop some basic geometric concepts in metric spaces which make precise certain intuitive ideas centered on the themes of \interior" and \boundary" of a subset of a metric space. This completes the proof. The above definitions (open ball, open set, closed set ...) all make sense when $n=1$, that is, for subsets of $\R$. 1 Interior, closure, and boundary Recall the de nitions of interior and closure from Homework #7. endobj /pgf@CA0.2 << /Contents 57 0 R This makes x a boundary point of E. /pgf@ca0.6 << /PaintType 2 The second $\iff$ follows directly from the definition of interior point. stream << • The complement of A is the set C(A) := R \ A. To do this, we must prove that $\forall \bfx\in S$, condition \eqref{interior} holds. De Morgan's laws state that $(A\cup B)^c = A^c \cap B^c$ and $(A\cap B)^c = A^c \cup B^c$. \begin{align} &\iff /Annots [ 56 0 R ] $$ endobj 5 | Closed Sets, Interior, Closure, Boundary 5.1 Deﬁnition. 12 0 obj an open interval $(a,b)$ is an open set. /FirstChar 27 $$ /pgf@ca.4 << In other words, A point that is in the interior of S is an interior point of S. $$ /Type /Catalog $$ We use d(A) to denote the derived set of A, that is theset of all accumulation points of A.This set is sometimes denoted by A0. /CA 0.2 /Encoding 22 0 R $$ concepts interior point, boundary point, exterior point , etc in connection with the curves, surfaces and solids of two and three dimensional space. /ca 0.5 >> A closed interval $[a,b]$ is a closed set. \begin{align} a set $S\subset \R^n$ can be neither open nor closed. [7]. /CA 0.5 /CapHeight 696 >> /Contents 79 0 R Equivalently, $\bar S = S^{int}\cup\partial S =$ Case 1 $\cup$ Case 3. /ca 1 &\iff \ << /pgf@CA0.3 << What about Case 2 above? /Length3 0 /F63 46 0 R On the other hand, the proof that (spoiler alert for example 1 below) the every point of an open ball is an interior point is fundamental, and you should understand it well. \partial S = \partial (S^c). $\partial S\subset T$: We already know that if $ |\bfx-\bfa|

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