Connected (idea)
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In [topology], an important [axiom] of [topological space]s.
[Definition] A topological space X is said to be disconnected if there are [nonempty] subsets A,B of X such that AnB={}, AuB=X, and A and B are both open in X. Since A and B are [complement]s in X, it follows that they must both be closed. So we could replace 'open' with 'closed' in the definition above. I will refer to A and B as a 'disconnection' of X (which isn't standard terminology). It's reasonably clear that the following is equivalent: [Lemma] X is disconnected iff it has a proper, nonempty subset that is both open and closed. [Definition] A topological space X is connected if it is not disconnected. A subset Y of a topological space is connected if Y is connected in the [subspace] topology inherited from X. This definition sounds a bit odd at first, but it does give the desired effect - that is, the spaces you would expected to be connected are, and vice versa. Having said that, for spaces which aren't well [separation axioms|separated] (ie. they don't satisfy many [separation axioms]), deciding whether a space is connected can be far from an [mathematical intuition|intuitive] matter. A more intuitive (and stronger) [concept] is that of [path connectedness]. The following are the basic results about connectedness, leading to a proof of the connectedness of R^{n}; they're all pretty intuitive, and all have pretty basic proofs. Since the definition of connectedness is in negative terms, stating that a space is not disconnected, all the main proofs are by [proof by contradiction|contradition]. [Theorem] X=[0,1] is connected. [Proof] Suppose not. Then there are subsets A, B with AuB=X, AnB={}, A,B both open in X. Then 1 is in A or 1 is in B; suppose it's in A. Then, B is nonempty, and bounded above by 1, so it has a supremum b. Now, if b is in B, then b>1 since 1 is in A, and everything above b is in A. But B is open, so B must contain a small open [interval] centred on b, which must contain points above b, which is a contradiction. And if b is in A, then A contains a small open interval centred on B. But every open interval centred on b contains points of B, since b is the supremum of B, and this is also a contradiction. So [0,1] is connected. [Proposition] The [continuous] [image] of a connected space is connected. [Proof] Let X,Y be topological spaces, X connected, f:X->Y continuous and onto. Suppose Y is disconnected; then there is a disconnection A,B of Y. Then f^{-1}(A), f^{-1}(B) forms a disconnection of X, contradicting connectedness of X. [Proposition] If X is a topological space, Z is a connected subset of X, and Y is a subset of X containing Z and contained in the closure of Z, then Y is connected. In particular, if Z is connected then the closure of Z is connected. [Proof] Suppose A,B is a disconnection of Y. Then AnZ, BnZ is a disconnection of X. [Proposition] Suppose (X_{i})_{i in I} is a collection of connected spaces such [Intersection](X_{i}) is nonempty. Then Y=[Union](X_{i}) is connected. [Proof] Suppose A,B is a disconnection of Y, and suppose x is in Intersection(X_{i}). Then either x is in A or x is in B; suppose x is in A. Also, B is nonempty, so there is some i such that X_{i} contains a point of B. Then X_{i}nA and X_{i}nB are nonempty, so they form a disconnection of X_{i}. [Corollory] R is connected. [Proof] Define f_{n}:[0,1]->R by f(x)=2n(x - 1/2), so that f_{n}([0,1]) = [-n,n]. Then [-n,n] is the continuous image of the connected space [0,1], and so is connected. And R is union of the sets [-n,n] for n in N which have nonempty intersection, so R is connected. [Proposition] If X,Y are connected spaces then XxY is connected. [Proof] Suppose A,B is a disconnection of XxY. Then there are points (x_{A}, y_{A}) in A, (x_{B}, y_{B}) in B. Let C be the subspace {(x_{A},y) in XxY : y in Y} and D be the space {(x, y_{B}) in X,Y : x in X}. Then C is clearly [homeomorphism|homeomorphic] to Y, D is clearly [homeomorphism|homeomorphic] to X, so both are connected. Moreover, the element (x_{A}, y_{B}) is in both spaces, so their union Z=CuD is connected. But ZnA, ZnB form a disconnection of Z, which is a contradition. [Corollory] R^{n} is connected. See also: [path-connected], [quasi-connected], [locally connected], [component] | Existing:
Non-Existing: |