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Is this logically valid? And if so, could someone please explain to me the name of the logical construct I'm performing (e.g. converse, contrapositive) as well as its mechanics? Please note that my training in mathemtical logic is basically non-existent.

If for any $\epsilon$ there exists some $\delta$ such that for all $x$, $a - \delta < x < a + \delta$ implies $l - \epsilon < f(x) < l + \epsilon$; then, for any $\delta$ there exists $\epsilon$ such that for all $f(x)$, $l - \epsilon < f(x) < l + \epsilon$ implies $a - \delta < x < a + \delta$.

Ultimately I'm aiming to show that if $\lim \limits_{x \to a} f(x) = l$, then $\lim \limits_{y \to l} f^{-1}(y) = a$.

user_hello1
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    It is not clear what you are asking. Why would the function $f$ have an inverse? Take $f(x) = l$ for all $x$, for example. – copper.hat Sep 13 '19 at 21:10
  • I'm supposing $f$ does have an inverse. – user_hello1 Sep 13 '19 at 21:17
  • You can't just arbitrarily reverse the quantifiers and reverse the implication. (As I pointed out, a constant function satisfies this second criterion.) It's certainly not true in general that a one-to-one continuous function must have a continuous inverse, so you're going to have to work harder than this — e.g., to show that the continuous function must be increasing or decreasing. – Ted Shifrin Sep 13 '19 at 22:01
  • This is not logically valid - you reversed the implication, which in general cannot be done. If every apple is red, that doesn't mean that every red thing is an apple. ;) – mz71 Sep 13 '19 at 22:02
  • for any δ there exists ϵ such that for all f(x), l−ϵ<f(x)<l+ϵ implies a−δ<x<a+δ" that just plain isn't true. Try the $\sin$ function for example. For any $\delta$ let $k$ be an integer so that $2k\pi > \delta$. Let $a=\frac \pi 2$ and $x = \frac \pi 2 + 2k\pi$ and $l =1$ and $\epsilon$ be anything. Then $1-\epsilon <\sin x < 1+\epsilon$ is always true but $a-\delta < x < a+\delta$ is not. – fleablood Sep 13 '19 at 23:54
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    If $f$ is bijective so that $f$ has an inverse and $f$ is continuous, then what you are trying to proved but you can't just reverse the conditionals to prove it. You will have to prove that for any $\epsilon > 0$ there is a $\delta$ so that $|a-x|< \delta$ implies $|f^{-1}(x)-l| < \epsilon$. – fleablood Sep 14 '19 at 00:01
  • Have a look at https://math.stackexchange.com/questions/68800/functions-which-are-continuous-but-not-bicontinuous. – Rob Arthan Sep 15 '19 at 15:23

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