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February 15th, 2017, 10:56 AM   #1
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state an invertible function that is not continuous:

State an invertible function that is not continuous.

Would this function work?

f(x) = x if x is not equal to -1 and 1

and f(-1) = 1 and f(1) = -1
since this function is still surjective and injective.
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Last edited by skipjack; February 15th, 2017 at 02:00 PM.
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February 15th, 2017, 11:24 AM   #2
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a much easier example is just

$f(x) = \dfrac 1 x$
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February 15th, 2017, 02:07 PM   #3
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With $f$(0) = 0.
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February 15th, 2017, 02:52 PM   #4
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Quote:
Originally Posted by skipjack View Post
With $f$(0) = 0.
or just not included in the domain

0 isn't in the range
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February 15th, 2017, 03:11 PM   #5
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No, because omitting 0 from the domain makes the function continuous.
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February 15th, 2017, 04:16 PM   #6
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Quote:
Originally Posted by romsek View Post
a much easier example is just

$f(x) = \dfrac 1 x$
This function is continuous.

A strategy for this is to pick your favorite continuous injective function whose domain and range are not $\mathbb{R}$. Now pick any value not in the domain, $x_0$, and another value not in the range, $y_0$ and specify that your new function takes the values of the continuous function but additionally that $f(x_0) = y_0$.

Use this strategy to come up with an explicit example.
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