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Hello to all, here's another problem the answer of which I'd like to check.
Let X be a non-empty infinite set, and U = {Φ} U {X\K : K is a finite subset of X} (Φ denotes the empty set) a topology on X. One needs to prove that the topological space (X, U) is connected and compact.
Now, it seems quite obvious that the space isn't connected, since there don't exist two non-empty disjoint open sets which cover the space (for any two chosen finite sets, the open sets of the form X\K in the topology will have an infinite number of elements in common).
Further on, to show that the space is compact, let C be an open cover for X. Then X = [tex]\cup[/tex] X\K = X\[tex]\cap[/tex]K (I didn't index the set operators for practical reasons). Now, it follows that the intersection of finite sets [tex]\cap[/tex]K is empty, and so it must contain at least two disjoint finite sets Ki and Kj. It we take these two sets, then (X\Ki)U(X\Kj) is a finite cover for X, so (X, U) is compact.
Let X be a non-empty infinite set, and U = {Φ} U {X\K : K is a finite subset of X} (Φ denotes the empty set) a topology on X. One needs to prove that the topological space (X, U) is connected and compact.
Now, it seems quite obvious that the space isn't connected, since there don't exist two non-empty disjoint open sets which cover the space (for any two chosen finite sets, the open sets of the form X\K in the topology will have an infinite number of elements in common).
Further on, to show that the space is compact, let C be an open cover for X. Then X = [tex]\cup[/tex] X\K = X\[tex]\cap[/tex]K (I didn't index the set operators for practical reasons). Now, it follows that the intersection of finite sets [tex]\cap[/tex]K is empty, and so it must contain at least two disjoint finite sets Ki and Kj. It we take these two sets, then (X\Ki)U(X\Kj) is a finite cover for X, so (X, U) is compact.