 Consider the projection on the ﬁrst variable p1: R2 → Rdeﬁned by p1(x,y) = x. /Length1 1543 It combines both simplicity and tremendous theoretical power. The intervals are precisely the connected subsets of {\displaystyle \mathbb {R} }. I'm new to R and I'm trying to get my script more efficient. �4U3I5��N�g�_��M�����ô:���Zo�N߽z?��A�A�pX����~L����n A (connected) component of a topological space is a maximal connected subset. Both are aleph_2 2 #R (see jhdwg's comment). Let us consider r, s. One can check the following observations: ∗ [r,s[ is bounded, ∗ ]r,s] is bounded, and ∗ ]r,s[ is bounded. Close Menu. \f 1(closed) = closed" 3. Focusing for the moment on the real line R, one uses the completeness property of the usual ordering to show that the connected subsets of R are the intervals; i.e., … Calculus and Beyond Homework Help. �Sm��N�z��ʾd�ƠV��KI�Bo{� ��� ����IL�w��o &d�*�z�֗d*H 2�\z��aʌ����d34:������ S*D0�Ǵ��JH�B�YB1�_l�B���%��A�i��R I_���R �3�X+S�'���G� R�w7������~@�}��Z First we need a lemma. Here is one thing to be cautious of though. However, the number of observations (lines) for each subject is not equal.I would like to separate my subjects into groups, according to their number.How can I do it? The continuous image of a connected space is connected. Every If is empty or has only one element, the required result holds, ... Let be an interval. proof: Let X R be a subset of R that is not an interval. If S is any connected subset of R then S must be some interval. (In particular, so are Rn itself, the ball Bn, and the disk Dn.) k are intervals, so m(I k) = l(I k) = m(I k). Pick any and in with . If X is an interval P is clearly true. A component of Q is a maximal connected subspace. The components of Q with the absolute value topology are the one-point subspaces. Solution to question 2 . Thus, f is continuous in that case as well. For simplicity's sake, we will only give the proof when the interval is closed, and without loss of generality the interval can be assumed to be [0, 1] [0,1] [0, 1] in this case. The complement of a subset Eof R is the set of all points in R which are not in E. It is denoted RnEor E˘. Curves are important geometric objects, especially in R 2 and R 3 , because they describe traces of particles when the variable x is interpreted as time. Let c2A\B. Consider the projection on the ﬁrst variable p1: R2 → Rdeﬁned by p1(x,y) = x. Then the only subsets of Y which are open are ∅,Y and their inverse images are ∅,X which are both open in X. A component of Q is a maximal connected subspace. Then the subsets A (-, x) and A (x, ) are open subsets in the subspace topology A which would disconnect A and we would have a contradiction. Actually, the Cantor set is totally disconnected (its only nonempty connected subsets are singletons) because, by a homework exercise, the Cantor set contains no intervals. As we saw in class, the only connected subsets of R are intervals, thus U is a union of pairwise disjoint open intervals. & ON CLOSED SUBSETS OF R AND OF R2 ADMITTING PEANO FUNCTIONS Abstract In this note we describe closed subsets of the real line P ˆ R for which there exists a continuous function from P onto P2, called Peano function. Let U be an open subset of R. As any set, U is a union of its connected components. 11.W. We will give a short proof soon (Corollary 2.12) using a different argument. %���� B ⁢ (x, r) = {y ∈ X | d ⁢ (x, y) < r}. © 2003-2020 Chegg Inc. All rights reserved. Let O subset R be open. Since intervals are connected by Theorem 8.30, part (b) let’s us conclude that E:= [[a;b]2I [a;b] (4) is connected. Suppose to the contrary that M is a component and contains at least two distinct rational numbers, p and q. Answer: I’ll start with the n = 1 case, so suppose that U is a nonempty open subset of R1, and assume that its complement is nonempty; I will show that U cannot be closed. The subset (1, 2) is a bounded interval, but it has no infimum or supremum in P, so it cannot be written in interval notation using elements of P. A poset is called locally finite if every bounded interval is finite. Let U be an open subset of R. As any set, U is a union of its connected components. Theorem 6. Thus f(U) will always be a subset of Y, and f 1(V) will always be a subset of X. 7 0 obj << Theorem 5. Any interval in R \mathbb R R is connected. \f(compact) = compact" 4. >> Then the only subsets of Y which are open are ∅,Y and their inverse images are ∅,X which are both open in X. Privacy Every interval in R is connected. be used only for subsets of Y. Before I used this formula: ����0�������@Rgst��]��υ.\��=b"��r�ġn Show that the topology on R whose basis is the set of half-open intervals [a, b) is normal. Proof that any non-interval is not connected: Let I be a subset of R that is not an interval. This should be very easy given the previous result. ? The notion of topological connectedness is one of the most beautiful in modern (i.e., set-based) mathematics. Then there is a point t =2I such that the two sets G 1 = (1 ;t) \I; G 2 = (t;1) \I are both non-empty. School Stanford University; Course Title MATH 171; Type. More options. X cannot be written as the union of two nonempty separated sets. | A T 1 space is one in which for every pair of points x y there is an open set containing x but not y. xڍ�P�.w-P,�������C A�w�R��)VܝRܵ@�k/�����3�Nf�}^�}^���R�k�JZ��Ar'WVN6!����>'���������V������V�ڂ����B�>�d��φ�'�������'��/�������� �m-�l %��J+ v���Z۸>�������g�� ���Z � �@W���@������_!Dl\]����=<�a�j�AAw�%�7e���56TZ���-�O������ '賋��%x����x�r��X�O�_�p�q��/�߁l��pZX���N^�N� +[൜ Solution to question 2 . Proof. We all know what intervals are from high school (and we studied the nine di erent types on homework 6). A set X ˆR is an interval exactly when it satis es the following property: P: If x < z < y and x 2X and y 2X then z 2X. In mathematics, a (real) interval is a set of real numbers that contains all real numbers lying between any two numbers of the set. Hint: Suppose A CR is nonempty and connected. An open subset of R is a subset E of R such that for every xin Ethere exists >0 such that B (x) is contained in E. For example, the open interval (2;5) is an open set. %PDF-1.5 A subset of the real line R that contains more than one point is connected if and only if it is an interval. By: Search Advanced search … Menu Log in Register Navigation. This theorem implies that (0;1) is connected, for example. Show that the only connected subsets of R are the empty set, singleton sets {a}, a E R, and 'intervals' of the following forms: (-0,00), (a,00), [a, 20), (-00, a),(-0, a], (a,b], [a,b), (a,b), [a, b], where a, b E R, and a < b when appropriate. Then The select argument exists only for the methods for data frames and matrices. the intervals in R are the nonempty connected subsets of the real line. We claim that E= A\B, which will nish the proof. For example, the integers are locally finite under their natural ordering. The union of open sets is an open set. Theorem 5. Proof. We prove that is connected: there do not exist non-empty open sets and in , such that and . Let (X, d) be a metric space. endstream For both proofs - the criterion of connectedness and the property of separated sets - one needs some basic topology, which I don't … C���Y�h6��#��u��~�/���Aee�b_UE1av�n{���F�&�0;1t��)��;������Ь"h8�O 5� �~ ��Z��,D��Z�����ύG�l/"ZqRB ���J���,wv��x�u��_��7 Proof. (iii) is an interval. \begin{align} \quad \delta = \min \{ \| \mathbf{x} - \mathbf{s_1} \|, \| \mathbf{x} - \mathbf{s_2} \|, ..., \| \mathbf{x} - \mathbf{s_n} \| \} \end{align} Show that the set A = {(x,y) ∈ R 2: x > 0} is open in R2. Open interval: all cut points Half-open interval: one non-cut point 1. The continuous image of a connected space is connected again. 1. Show that the only connected subsets of R are the empty set, singleton sets {a}, a E R, and 'intervals' of the following forms: (-0,00), (a,00), [a, 20), (-00, a), (-0, a], (a,b], [a,b), (a,b), [a, b], where a, b E R, and a < b when appropriate. Moreover, Q is not locally connected. If 5 C R is not an interval, there exist a,b £ 5 and c ^ S such that a < c < b. The range of a continuous real function defined on a connected space is an interval. (20 Points) We Proved In Lecture That The Only Connected Sets In R Are The Intervals. Proof: If S is not an interval, then there exists a, b S and a point t between a and b such that t is not in S. Then define the two sets U = ( - , t ) and V = ( t, ) Then U S # 0 (because it contains { a }) and V S # 0 (because it contains { b }), and clearly (U S) (V S) = 0. Both G 1 and G 2 are open in the subspace topology (by de nition), they are non-intersecting, and I = G 1 [G 2. Problem 11: Prove that if a ˙-algebra of subsets of R contains intervals of the form (a;1), then it contains all intervals. A set is clopen if and only if its boundary is empty. The proof for connectedness I know uses the theorem, that the empty set and the entire space are the only subsets of a connected space, which are open and closed - and vice versa. Suppose to the contrary that M is a component and contains at least two distinct rational numbers, p and q. Recall that for x ∈ X and r ∈ ℝ + we have. Path-connectedness. We have shown that connected sets in R must be intervals. Each closed -nhbd is a closed subset of X. Theorem 4. In particular, an image of the closed unit interval [0,1] (sometimes called an arc or a path) is connected. As It Turns Out, Connectedness Of A Set Is Equivalent To The Claim That The Only Simultaneously Closed And Open Subsets Of It Are Itself And The Empty Set. We first prove that (i) implies (ii). (2) To see that Rx>0 for all x, suppose there is an x where Rx=0. In particular, an image of the closed unit interval [0,1] (sometimes called an arc or a path) is connected. Let be connected. 11.V Corollary. Lemma 1. If all connected subsets of a real number set are intervals only, then what about the null set and singleton sets? Any open interval is an open set. We wish to show that intervals (with standard topology) are connected. Subsets of the real line R are connected if and only if they are path-connected; these subsets are the intervals of R. Also, open subsets of Rn or Cn are connected if and only if they are path-connected. (This fact will not adapt if we were doing rectangles in R2 or boxes in Rn, however.) A T 1 space is one in which for every pair of points x y there is an open set containing x but not y. /Filter /FlateDecode The connected subsets ofR are precisely the intervals (open, half- open, or closed; bounded or unbounded). It follows that the image of an interval by any continuous function is also an interval. In order to this, we will prove that the space of real numbers ℝ is connected. So suppose X is a set that satis es P. Let a = inf(X);b = sup(X). 5. Since U is open, these connected components are open by Exercise 11. Actually, the Cantor set is totally disconnected (its only nonempty connected subsets are singletons) because, by a homework exercise, the Cantor set contains no intervals. Additionally, connectedness and path-connectedness are the same for finite topological spaces. 9.4 (3) Proposition. Prove that in Rn, the only sets which are both open and closed are the empty set and all of Rn. 11.U. Indeed, from the de nition of Ewe have that EˆA\Bsince each interval on the right hand side of (4) is assumed to be a subset A\B. Let us consider r, s. One can verify the following observations: ∗ [r,s] is connected, ∗ [r,s[ is connected, ∗ ]r,s] is connected, and ∗ ]r,s[ is connected. First we need a lemma. Proof. A poset can only have one greatest or least element. We will give a short proof soon (Corollary 2.12) using a different argument. It is not very hard, using theGG‘ iff least upper bound property of , to prove that every interval in is connected. Proof: If is empty or has only one element, the required result holds, so we may assume that has at least two elements. >> Homework Help . Then as separated sets ##X,Y## are both open and closed in ##X\cup Y = U(c,r)##. If all connected subsets of a real number set are intervals only, then what about the null set and singleton sets? Every B ⁢ (x, r) = {y ∈ X | d ⁢ (x, y) < r}. intervals are connected. Intervals in R1 are connected. Our characterization of those sets is based on the number of connected components of P. We also include a few remarks on com-pact subsets of R2 admitting Peano … Theorem: The only connected subspaces of R are the intervals. Properties that are preserved in one direction or the other First, short-hand names to help you remember which facts are true. (1) For x in O, let Rx=inf{ r>=0 : N(x,Rx) contained in O}, where N(x,e) is the interval of radius e centered at x. The only continuous functions from X to {0,1}, the two-point space endowed with the discrete topology, are constant. Theorem 4. show any interval in R is connected. So we assume given a real interval and subsets and of each non-empty and each open in , such that and . See my answer to this old MO question " Can you explicitly write R 2 as a disjoint union of two totally path disconnected sets?". Search titles only. Intervals In the sequel a, b, r, s are real numbers. Then is not a subset … 29 0 obj << \f 1(open) = open" 2. Since U is open, these connected components are open by Exercise 11. There are locally connected subsets of\mathbb{R}^2$which are totally path disconnected. School Stanford University; Course Title MATH 171; Type. T6–3. A subset of the real line R that contains more than one point is connected if and only if it is an interval. Solution: Let be the ˙-algebra. Prove that the only T 1 topology on a finite set is the discrete topology. The range of a continuous real function defined on a connected space is an interval. stream Homework Help. Each convex set in Rn is connected. We claim that E= A\B, which will nish the proof. /Length3 0 Both R and the empty set are open. Both G 1 and G 2 are open in the subspace topology (by de nition), they are non-intersecting, and I = G 1 [G 2. Let us consider r, s. One can verify the following observations: ∗ [r,s] is connected, ∗ [r,s[ is connected, ∗ ]r,s] is connected, and ∗ ]r,s[ is connected. We must show that if and are in , with , then . Theorem 6. (A is not a clopen subset of the real line R; it is neither open nor closed in R.) Properties. Since intervals are connected by Theorem 8.30, part (b) let’s us conclude that E:= [[a;b]2I [a;b] (4) is connected. T6–3. In fact, a subset of is connected is an interval. /Length 2688 So combining both the theorems we conclude that a subset of R is Connected if and only if it is an interval. Ԏ��b��>�� ���w3�F����k������6���"9��>6��0�)0� �)�=z�ᔚ�v ��Df��W�>^e�Z�Ң��#_���o d�O.Yެ]S��Z�in��. 5), and Si and 52 are nonempty since a €. Finally we proved that the only connected bounded subsets of R are the empty. 3. Open interval: all cut points Half-open interval: one non-cut point 1. Connected subset Thread starter tarheelborn; Start date Oct 19, 2010; Oct 19, 2010 #1 tarheelborn. Solution. (If you can’t ﬁgure this out in general, try to do it when n = 1.) intervals ( 1;1=2) and (1=2;2), each of which has nonempty intersection with the Cantor set. �f1ٰlg�-7;�����GQrIN!&�?�i�, ���*�t�H4��.S���ӣ�Ys�3�N# �X ��Ŷ*�ò~�W3�|���c HS��oI_�f;.�&����E�ڢG�,��q>^���is����B ����v�l��C�'��O��䝂�Lnl��>EՂ ��H%4�Ao��o����}�>N��L�"�͉��t�U�݃��æ�2)����J�芈���Lmrs9{Ю����sH�Q'!8ήF?Ds�$���z}(Q4j,�������bSl�L*���X��wXlk��!,���V�H+RH2�6: $G��>��w���rL��Y��@oC�aN�@5A-�2�GҪ ��W۹�47��x@H�8��l��$ce?��,��=:]r�-�ã���� {�/�d� ���7j��0 J�Q�@EF92��b�&c[�ʵX��b��U���PrhkQʩHѧǠ)1qb!��_:L�� �/ؾ(�+n��%�� &�bM�)�t�c�=|J^�߹'����e�T]_�\�릐K(���L�dF�b���h�B;�-��GL��y�(N�av���G+,��U�m��y���L������vwn��ak�E�lY��x�׶G�5�_�Y-�����аxwqg)Tڳ��Y�.�ȡ��u�Wyf�y�e����ݹ*!�F�0���7�@��QRau�����P&�O�t�9Ζ�X|r�����(w��#�>������ b�������v��8�[z��l�����:�P*���9R����L{ Every star-shaped set in Rn is connected. stream endobj ���w,��w��� _6-�"��h�@i E�s��g��E��0�f�ߜ���mc��Z Օ]u.d+�q��a%�Wz___/R�0�R���s����x,!&��{"R葡��aF� Forums. We allow a = 1 ;b = +1. This is one formulation of the intermediate value theorem. Proof. Mathematics 468 Homework 2 solutions 1. Hint: Suppose A CR is nonempty and connected. The question can be rephrased as “ Can the null set and singleton sets be connected sets? Terms The question can be rephrased as “ Can the null set and singleton sets be connected sets? (‘‘Try it as an exercise!) #&�Q��DE���΁s�ցu0���c�G�p�i�b��Ԛ�xL�b�:�]��R�Q,�y�X�A�� c�$�T In Particular This Proves That The Set R Itself Is Connected. A topological space X is connected if and only if the only clopen sets are the empty set and X. Contact us. Prove that a space is T 1 if and only if every singleton set {x} is closed. Prove that the intersection of connected sets in R is connected. Finally we proved that the only connected bounded. The connected subsets of R are exactly intervals or points. Indeed, from the de nition of Ewe have that EˆA\Bsince each interval on the right hand side of (4) is assumed to be a subset A\B. A subset of a line is connected iﬀ it is an interval. Suppose that is not a subset of . Show that this is false if “R” is replaced by “R2.” Proof. Feel free to say things like this case is similar to the previous one'a lot, if it is actually similar... TO Proof. The precise versions are given after the list. When you think about (0;1) you may think it is not Dedekind complete, since (0;1) is bounded in R and yet has no upper bound in (0;1). * Prove that every connected subset of R is an interval. The connected subsets ofR are precisely the intervals (open, half-open, or closed; bounded or unbounded). By complement must contain intervals of the form (1 ;a]. Divide into a bunch of cases, e.g. intervals are connected. As we saw in class, the only connected subsets of R are intervals, thus U is a union of pairwise disjoint open intervals. Then 5 = Si U 52 (since c fi 5), and … Prove that any pathwise connected subset of R(real numbers) is an interval. /Length 10382 Show that the set A = {(x,y) ∈ R 2: x > 0} is open in R2. View desktop site. Moreover, Q is not locally connected. Proof. We rst discuss intervals. Intervals In the sequel a, b, r, s are real numbers. Path-connectedness. Let Si = S n (-00, c) and 52 = 5 n (c, 00). 11.X Connectedness on Line. Pages 3. Proof. Proof: We assume the contrary and derive a contradiction. �\ Ͼ�W᎚�l>]���]��;6S���Ԁ*bw��t�#�ܙF��P�Լ�����rFH�ٳ*[V�E���{�3 Pages 3. 3. (‘‘Try it as an exercise!) For the second, you can map R 2 to a disk in another R 2 and draw a circle enclosing the cone, touching it at the vertex. Then 5 = Si U 52 (since c. fi. Click for a proof All proofs of this result use some form of the completeness property of R. \mathbb R. R. Here is one such proof. It works by first replacing column names in the selection expression with the corresponding column numbers in the data frame and then using the resulting integer vector to index the columns. Let (X, d) be a metric space. If 5 C R is not an interval, there exist a,b £ 5 and c ^ S such that a < c < b. 1 MeasureTheory The sets whose measure we can deﬁne by virtue of the preceding ideas we will call measurable sets; we do this without intending to imply that it is not possible Any clopen set is a union of (possibly infinitely many) connected components. Prove that a space is T 1 if and only if every singleton set {x} is closed. Prove that the only T 1 topology on a finite set is the discrete topology. Homework Help . * Seperated sets, connected sets in metric space - definition and examples. Divide into a bunch of cases, e.g. (^H>�TX�QP����,9I�]^]m���e�� r8���g3��"� ��EI'Qb���[�b�q7'�N��| �\}�*����D�8��!NH�� Q�\ �ޭ��~\�9.F6Y�8ށ��L =l��)�K6��t����d�H�.���mX��S��g��{�|^� ���ޯ�a W�:b�� �?������vu�B��6E(:�}� �r���B����0�T�IK���ve�x�2�ev��@И�#�w"۽��@�:11«����*�-O/��zp�S:���4����l��I�5Td'�����4�Ft;�?���­ZԿeQW�� �֛U6�C���29�yx�W*���.zއ���؀� d� x��[[o�~��P�Fh��~�n�X/�6A����@�E�l����8������| �k$Q��wn�9d�����q�'^�O�^�!rF�D���Ō Let Si = S n (-00, c) and 52 = 5 n (c, 00). 9.4 (3) Proposition. De nition 5.22. In order to this, we will prove that the space of real numbers ℝ is connected. This preview shows page 2 - 3 out of 3 pages. A similar proof shows that any interval is a connected subset of R. In fact, we have: Theorem Intervals are the only connected subsets of R with the usual topology. This was answered by the next theorem. Theorem 2.7. Let us consider r, s. One can check the following observations: ∗ [r,s[ is bounded, ∗ ]r,s] is bounded, and ∗ ]r,s[ is bounded. Proof If A R is not an interval, then choose x R - A which is not a bound of A. Finally we proved that the only connected bounded. "N�I�t���/7��Պ�QOa�����A����~�X��Ə߷fv��@�Wۻ��KЬ3��Sp�����3)�X!Au���?�6���f?�5�^��%)ܩ��H]��_�Y�$����Bf��9Ϫ�U��FF�R�#hVPQ�߳�c�!�t���H��ʲ����#*�}�#4{�4i�F��7���D�N����H��b��i�aubT+��{ȘNc��%�A��^&>�5��$xE��2.����;�ʰ�~w[����ɓ��v���ۛ9��� ��M��4�J����@ ^-�\6"z�.�!h��J�ᙘQ������}��T��+�n�2?c�O�}�Xo.�x=���z� Yd�ɲ����ûA�=HU}. /Filter /FlateDecode Fur-thermore, the intersection of intervals is an interval (possibly empty). Let and . Prove that the only connected subsets of R are (a) the empty set, (b) sets consisting of a single point, and (c) intervals (open, closed, half-open, or infinite By Theorems 5.24 and 5.31 Theorem 5.24 Theorem 5.31, the curve C is a compact and connected subset of R k since it is a continuous image of a compact and connected set. ���+ �d��� ?�݁�@�g�?��Ij �������:�B٠��9���fY'Ki��#�����|2���s޽��*������ode�di�����3�����HQ�/�g�2k�+������O r��C��[�������z��=��zC�� �+� ������F��� K[W�9��� ����b�՟���[O�!�s�q8~�Y?w�%����_�?J�.���������RRO�7+/���������^��w�2�7�?��@ۿN���� �?I. The only subsets of X with empty boundary are X and the empty set. Uploaded By ruijiestanford. /Length2 9365 !,u~�6�M\&T���u-���X>DL�Z ��_̶tb������[F!9����.�{�f��8��Ո��?fS?��n�1DY�R��P1�(�� �B���~ʋ���/g ��� The continuous image of a connected space is connected again. Connected Sets in R. October 9, 2013 Theorem 1. If S is any connected subset of R then S must be some interval. Show that the only connected subsets of R are the empty set, singleton sets {a}, a E R, and 'intervals' of the following forms: (-0,00), (a,00), [a, 20), (-00, a),(-0, a], (a,b], [a,b), (a,b), [a, b], where a, b E R, and a < b when appropriate. 11.T. Hence, as with open and closed sets, one of these two groups of sets are easy: open sets in R are the union of disjoint open intervals connected sets in R are intervals The other group is the complicated one: closed sets are more difficult than open sets (e.g. 1. This preview shows page 2 - 3 out of 3 pages. Proof that any non-interval is not connected: Let I be a subset of R that is not an interval. Prove that R1 is connected. We wish to show that intervals (with standard topology) are connected. Thus, f is continuous in that case as well. Lemma. Show that the topology on R whose basis is the set of half-open intervals [a, b) is normal. Recall that for x ∈ X and r ∈ ℝ + we have. Uploaded By ruijiestanford. Continuous images of connected sets are connected. The set {x in R | x d } is a closed subset of C. Each singleton set {x} is a closed subset of X. Let AˆR be a subset of R. Then x2R is: (1) an interior point of Aif there exists >0 such that A˙(x ;x+ ); (2) an isolated point of Aif x2Aand there exists >0 such that xis the only point in Athat belongs to the interval (x ;x+ ); (3) a boundary point of Aif for every >0 the interval (x ;x+ ) contains A (connected) component of a topological space is a maximal connected subset. In fact, a subset of is connected is an interval. The components of Q with the absolute value topology are the one-point subspaces. Continuous images of connected sets are connected. Exercise. 2 Hint: Suppose A CR is nonempty and connected. intervals ( 1;1=2) and (1=2;2), each of which has nonempty intersection with the Cantor set. Then there is a point t =2I such that the two sets G 1 = (1 ;t) \I; G 2 = (t;1) \I are both non-empty. Finally we proved that the only connected bounded subsets of R are the empty. It is not very hard, using theGG‘ iff least upper bound property of , to prove that every interval in is connected. Solution. Let c2A\B. A subspace of R is connected if and only if it is an interval. The cardinality of all subsets of R is aleph_2 2 #R (see jhdwg's comment), and you can go from a subset of R to a connected subset of R 2 (with R included as the x-axis) by connecting each point to (0,1). A is bounde above or not, and if it is bounded above, whether sup A E A or not. Professor Smith posed the question \Are there subsets of R that are connected but not one of the 9 intervals discussed?" Only continuous functions from x to { 0,1 }, the only clopen sets are empty. R } ^2 $which are totally path disconnected from x to 0,1...$ which are both open and closed are the empty these connected.! M ( I k ) = { ( x, suppose there is an interval p clearly... 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