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Actually, I have found some proofs to this but I have not tackled about compact, Heine-Borel theorem, metric spaces, sequences and series, etc. in my class..

- Thread starter irony of truth
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- #1

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Actually, I have found some proofs to this but I have not tackled about compact, Heine-Borel theorem, metric spaces, sequences and series, etc. in my class..

- #2

TD

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The proof I have uses Heine-Borel, so I assume you won't be interested in that one.

I do doubt thought that you would be able to rigourously proof it without any of the concepts or theorems you mentioned, unless you're just interested in explaining it, the difference between 'normal' and 'uniform' continuity.

I do doubt thought that you would be able to rigourously proof it without any of the concepts or theorems you mentioned, unless you're just interested in explaining it, the difference between 'normal' and 'uniform' continuity.

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matt grime

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Thank you for the help...

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mathwonk

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Then ther is also an infinite num=ber of them between say .11 and .12.

and so on. Eventually (after an infinite number of steps) you construct an infinite decimkal, i.e. a real number, with an infinite number of these points in every neighborhood.

Since the infinite sequence was chosen in the beginning to be points hwere the function was "less and less continuous" i.e. needing bigger and bigger delta for a given epsilon, you deduce that your function is not continuous at all at the coinstructed point. a contradiction.

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