A topology T is defined on the real line R by T = {,R}{(a,a)|aR and a>1}. (a) De

Question

A topology T is defined on the real line R by
T = {,R}{(a,a)|aR and a>1}.

(a) Describe the collection of closed sets in this topology on R.

(b) Show that the function f : (R,Teuclid) (R,T ) given by f(x) = x2 for x R is continuous.

(c) Show that the function g : (R, T ) (R, Teuclid) given by g(x) = x2 for x R is not continuous.

(d) Find all real numbers x for which the sequence (1/n) = (1, 1/2, 1/3, . . .) converges to x in this topology T .

Definition. A sequence (xn) = (x1, x2, x3, . . .) in a topological space converges to x if for every neighborhood U of x there is N Z+ such that xn U for all integers n N.

Explanation / Answer

(a) Since T is defined as a topology (as opposed to being a topology generated by T), we can assume all of the nontrivial (i.e. not R or ) open sets of T have the form (-a, a), where a is a real number and a > 1. Thus, the closed sets of T have the form R - (-a, a) = (-,-a] U [a, ), where a>1 is a real number. Oh yeah, R - R = and R - = R are closed sets too, but they always are in any topology. (note: you can PROVE that T is actually a topology, but since you didn't ask, I didn't show).

(b) To show f is continuous, we need to show the inverse image of an open set in T is also an open set in Teuclid (which I assume is the standard topology inherited from the standard Euclidean distance formula, i.e. the topology generated by all open intervals). So, take any nontrivial open set (-a, a), a >1, in T. Then f-1[(-a, a)] is the set of all real numbers x such that -a < x2 < a. It should be clear that this set is (-a, a), which is open in Teuclid (because if a > 1, then a > 1 too). As for R and , both f-1[R] = R and f-1[] = are open in T too. Therefore, f is continuous.

(c) To show g is not continuous, we only need to find one open set in Teuclid whose inverse image is not an open set in T. Consider the open set (1, 4) in Teuclid. Then g-1[(1, 4)] is the set of all real numbers x such that 1 < x2 < 4. It should be clear that this set is (-2, -1) U (1, 2). However, this set is not open in T, since it is not of the form (-a, a). Therefore, g is not continuous.

(d) Crazily enough, the sequence "converges" to any real number under the T topology. Why? Choose any real number x. A nontrivial open set that contains x must be of the form (-a, a), where a > |x|. But this interval always contains EVERY element of your sequence, since a > 1. Therefore, your sequence converges to x. Note: If you consider the trivial open set R as the one that contains x, then obviously it contains all of the sequence too.

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