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Unformatted text preview: MAT 325: Topology Professor Zoltan Szabo Problem Set 4 Rik Sengupta rsengupt@princeton.edu February 28, 2010 1. Munkres, p. 152, problem 5 A space is totally disconnected if its only connected subspaces are onepoint sets. Show that if X has the discrete topology, then X is totally disconnected. Does the converse hold? Solution. Suppose X has the discrete topology. Then consider a subset S X , where S contains more than one point. The claim here is that S is disconnected. To see this, suppose U = { s } for some s S , and take V = S { x } . Now, V is obviously nonempty, because we defined S as a subset with more than one point . But we know that singletons and (arbitrary) unions of singletons are open in the discrete topology (by definition, as in the discrete topology, all subsets are open, i.e. singleton sets form a basis for this topology). It follows that both U and V are open. Also, it is clear from the way we defined them, that U V = , and U V = S . Therefore, U and V , in fact, form a separation of S , and hence, S is disconnected, as claimed. But S was an arbitrary subset with more than one element, and it follows that all subspaces of X with more than one point are disconnected. Furthermore, it is obvious that one point sets are (trivially) connected. Hence, X is totally disconnected, and we are done. The converse does not hold. One example is the closure K of the set K = { 1 /n : n N } , with the standard subspace topology inherited from R . Then consider any subset S K that contains more than one point. Then, clearly, it has at least one point of the form 1 /n for some n N . This is because the closure of K has K itself and the point 0, so any subset with more than one point can at most contain zero and some point of K , i.e. it has to contain at least one point from K . But then, in the subspace topology, U = { 1 /n } is clearly open, closed, nonempty, and a proper subset of S , and hence, S is not connected. But the onepoint set { } is not open, so the space is not discrete. A somewhat nicer counterexample is the space Q R , with the subspace topology. First of all, notice that Q is totally disconnected. To see this, consider any subset S Q where S contains at least two distinct points q 1 and q 2 . But we know that the irrationals numbers are dense in R (A simple proof goes as follows: we know rationals are dense in R . Take two reals a and b with a < b . Then, we know, a 2 < b 2 , so we can find a nonzero rational number r between them. Then, we get a < r 2 < b , where r Q = r 2 R Q , thereby completing the proof). So, we can find some x R Q , such that q 1 < x < q 2 . Then, consider U = S ( ,x ), and V = S ( x, ). These two sets clearly give a separation of S , which is therefore disconnected. In other words, any subspace of Q containing more than one point is disconnected, and therefore, Q is totally disconnected, as claimed....
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This note was uploaded on 05/07/2010 for the course MAT 325 taught by Professor Zoltánszabó during the Fall '09 term at Princeton.
 Fall '09
 ZoltánSzabó
 Topology, Sets

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