caledonia Posted April 17, 2014 Share Posted April 17, 2014 From the usual epsilon, delta definition of continuity, two significant theorems follow : (i) a function continuous on a closed interval attains in supremum and infimum on that interval ; and (ii) for any two values in the range of the function, any intermediate value also belongs to the range. From these theorems, one can deduce that the image of any sub-interval of the domain is a sub-interval of the range. And it turns out that this is a necessary and sufficient condition for continuity. So we can use this last as the very definition of continuity for a function on a closed interval. Some students, who find the conventional definition of 'continuity at a point' disconcerting, may find this preferable. Link to comment Share on other sites More sharing options...

Orodruin Posted April 30, 2014 Share Posted April 30, 2014 What you suggest sounds very similar to the actual topological definition of continuity, for which it is not even necessary to have a metric space where epsilons and deltas have a meaning. As a student I admit to have been in awe of the beauty of topology as it allowed much more elegant proofs of many theorems. However, it is relatively abstract and I am not sure it is for everyone. Link to comment Share on other sites More sharing options...

Someguy1 Posted May 28, 2014 Share Posted May 28, 2014 (edited) From the usual epsilon, delta definition of continuity, two significant theorems follow : (i) a function continuous on a closed interval attains in supremum and infimum on that interval ; and (ii) for any two values in the range of the function, any intermediate value also belongs to the range. From these theorems, one can deduce that the image of any sub-interval of the domain is a sub-interval of the range. And it turns out that this is a necessary and sufficient condition for continuity. So we can use this last as the very definition of continuity for a function on a closed interval. Some students, who find the conventional definition of 'continuity at a point' disconcerting, may find this preferable. It's true that a continuous function takes closed intervals to closed intervals (as long as you allow a single point to count as a degenerate interval, which is reasonable]. A continuous function does not necessarily take open intervals to open intervals. For example f(x) = x^2 takes (-1, 1) to [0, 1). As far as the converses, it's false that a function that takes open intervals to open intervals is necessarily continuous. The counterexamples are pathological. I'm not sure whether a function that takes closed intervals to closed intervals must be continuous. Do you have a proof? Edited May 28, 2014 by Someguy1 Link to comment Share on other sites More sharing options...

caledonia Posted June 1, 2014 Author Share Posted June 1, 2014 It's true that a continuous function takes closed intervals to closed intervals (as long as you allow a single point to count as a degenerate interval, which is reasonable]. A continuous function does not necessarily take open intervals to open intervals. For example f(x) = x^2 takes (-1, 1) to [0, 1). As far as the converses, it's false that a function that takes open intervals to open intervals is necessarily continuous. The counterexamples are pathological. I'm not sure whether a function that takes closed intervals to closed intervals must be continuous. Do you have a proof? No I don't. Nor can I find a counter-example for closed intervals. Hence unknown whether my assertion is true or not ! Link to comment Share on other sites More sharing options...

caledonia Posted June 26, 2014 Author Share Posted June 26, 2014 I was wrong in my original post - it is not true that a function which maps closed (sub) intervals to closed (sub) intervals is necessarily continuous. Here is a counter example: Define f(x) as sin(1/(0.5-x)) for 0 <= x < 0.5, f(0.5) = 0, f(x) = sin(1/(x-0.5)) for 0.5 < x <= 1. Then f is discontinuous at 0.5, because for all delta > 0, there exists points nearer to 0.5 whose values are 1. But the image of any subinterval containing the point 0.5 is [-1,1] ; and the image of any subinterval not containing 0.5 is also a closed interval, because the function is here continuous and the intermediate value theorem applies. Link to comment Share on other sites More sharing options...

Someguy1 Posted June 26, 2014 Share Posted June 26, 2014 (edited) I was wrong in my original post - it is not true that a function which maps closed (sub) intervals to closed (sub) intervals is necessarily continuous. Here is a counter example: Define f(x) as sin(1/(0.5-x)) for 0 <= x < 0.5, f(0.5) = 0, f(x) = sin(1/(x-0.5)) for 0.5 < x <= 1. Then f is discontinuous at 0.5, because for all delta > 0, there exists points nearer to 0.5 whose values are 1. But the image of any subinterval containing the point 0.5 is [-1,1] ; and the image of any subinterval not containing 0.5 is also a closed interval, because the function is here continuous and the intermediate value theorem applies. I don't think this works. f is a continuous function. It's not defined at 0.5. At any point of its domain it satisfies the definition of continuity. By way of illustration, the function f(x) = x for x not equal 1/2; and f(1/2) = 47; is discontinuous at x = 1/2. But the function defined by f(x) = x on the domain R - {1/2} and undefined at x = 1/2; is continuous. The point is that we can't even ask about the continuity of my function or yours at x = 1/2, because it's not defined there. Edited June 26, 2014 by Someguy1 Link to comment Share on other sites More sharing options...

caledonia Posted June 27, 2014 Author Share Posted June 27, 2014 It is defined at x = 1/2, namely f(1/2) = 0. I used 0.5 instead of the symbol 1/2 ! Link to comment Share on other sites More sharing options...

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