Consider the set KCC consisting of the points of the form x+isin(1) for x > 0...
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Consider the set KCC consisting of the points of the form x+isin(1) for x > 0 together with the interval of points it with |t| ≤ 1. Show that K is closed and connected but not path connected. Connected means that if KC AUB with A and B open sets, then either KCA or KC B or AB is not empty. To show K is not path connected, try to show there is no continuous function (let alone piecewise differentiable) y with (0) = 0 and y(1) = 1 + i sin(1). To show K is connected, suppose K is contained in a union of two disjoint open sets. One of those sets must contain the interval {it} because the interval is connected. One of the sets must contain the graph {x+isin(1)} for the same reason. Consider the set KCC consisting of the points of the form x+isin(1) for x > 0 together with the interval of points it with |t| ≤ 1. Show that K is closed and connected but not path connected. Connected means that if KC AUB with A and B open sets, then either KCA or KC B or AB is not empty. To show K is not path connected, try to show there is no continuous function (let alone piecewise differentiable) y with (0) = 0 and y(1) = 1 + i sin(1). To show K is connected, suppose K is contained in a union of two disjoint open sets. One of those sets must contain the interval {it} because the interval is connected. One of the sets must contain the graph {x+isin(1)} for the same reason.
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The Kshow has closed Remember that the set of points of the type is what is described as the set K sin xisint for 0 x 0 in addition to the interval of ... View the full answer
Related Book For
Income Tax Fundamentals 2013
ISBN: 9781285586618
31st Edition
Authors: Gerald E. Whittenburg, Martha Altus Buller, Steven L Gill
Posted Date:
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