By E. M. Friedlander, M. R. Stein

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Ukm } of K. But then V ≡ Vk1 ∩ Vk2 ∩ · · · ∩ Vkm is an open neighborhood of x and U ≡ Uk1 ∪ Uk2 ∪ · · · ∪ Ukm is an open neighborhood of K, and V and U separate x and K. 9 A compact set in a Hausdorff space is closed. Proof: Let K be a compact set and x a point that is not in K. By the preceding proposition, there is a neighborhood U of x that is disjoint from K. That shows that the complement of K is open. So K is closed. 10 The hypothesis of “Hausdorff” is definitely needed in this last proposition.

And so does the point (2/π, 1). Suppose that γ is a continuous path-connecting the two points. We may take it that γ(0) = (0, 0). But then there are points t arbitrarily closed to 0 (of the form 2/[(2k + 1)π]) at which the function sin x1 takes the values ±1. So γ cannot be continuous. 8. 4 Let (X, U) be a topological space. If X is path-connected, then X is connected. Proof: Suppose to the contrary that X is disconnected. So there are disjoint open sets U, V that disconnect X. Let P be a point of U ∩ X and Q be a point of V ∩ X and γ : [0, 1] → X a path that connects them.

It is connected. For certainly the left-hand portion of S, which is the yaxis, is connected. And any open set that contains that portion will contain a neighborhood of the origin and hence intersect the right-hand portion (which gets arbitrarily close to the origin). 6 (The Intermediate Value Property) Let [a, b] be a closed, bounded interval in R. Let f be a continuous, real-valued function on [a, b]. Let γ be a real number that lies between f(a) and f(b). Then there is a number c between a and b such that f(c) = γ.