Proof for Images of Sets

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  179721 Images of Sets College level Proof before Real Analysis. Please see the attachment. Proof regarding one-to-one sets.

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  Different elements in are mapped to different images in under the function . Cartesian Product and Finite Sets and Algebra of Compositions of Functions (10 Problems) are investigated. The solution is detailed and well presented.

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  images of compact sets are compact), and finally, since N is Hausdorff, it follows that f(A) is closed, (since compact subsets of Hausdorff spaces are closed) This is a proof regarding bijection and homemorphism.

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  Proof:3 If the images of closed sets of X under f are closed in Y, this will prove continuity of the map f-1. If A is closed in X, then A is compact,then by proof2 , f(A) is compact. Since Y is Hausdroff, f(A) is closed in Y.

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  We know that f(S) denotes the set of the images of all elements of S under the function f.

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  Conversely, if , then for any , then for any , then . Thus . Hence we get . Therefore, This is a proof regarding intersection and union of sets.

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  Note that the same proof will not necessarily work for the intersection of infinitely many sets since we could not be sure that min { 1 , 2 , 3 , ...} > 0.

• Prove that the functions are subsets.

  d) E = f(f^-1(E)) if E is a subset of Rng(f) From Images of Sets. Prove for each part except (a) and (d) and explain each step of your solution. Please see the attachment.