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I feel it is, but cannot prove why. Also is it bijective, and is its inverse continuous?

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  • $\begingroup$ I've just drawn out a diagram, translating the points to their inverses and it looks right to me. I'm not sure how to begin testing with open sets or epsilon/delta $\endgroup$ – 123454321 Nov 20 '14 at 15:13
  • $\begingroup$ It's easy to do with $\epsilon/\delta$, go with that instead of open sets. $\endgroup$ – Henrik supports the community Nov 20 '14 at 15:17
  • $\begingroup$ If there were a point of discontinuity, what would it be? And what is the inverse function of $f(x)=1/x$? $\endgroup$ – Lubin Nov 20 '14 at 15:42
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Elaborating on that hint:

Let $p$ be an arbitrary point in $(1,\infty)$; you want to show the map is continuous at $p$. Fixate on some value of $\delta > 0$. Now look at the region near $1/p$. For some $\epsilon_1 > 0$, which will depend on both $p$ and $\delta$, the entire interval $(x-\epsilon_1,x)$ maps into the interval $p,p+\delta$ -- find the largest such $\epsilon$ explicitly, by solving $f(x) = p$. (This works in for this function because it is monotonic.)

Similarly, find $\epsilon_2$ such that the interval $(x,x+\epsilon_2)$ maps into the interval $p-\delta,p$ . Then let $\epsilon$ be the smaller of those two $\epsilon_i$; this $\epsilon (p,\delta)$, used in the continuity definition, shows that the function is continuous at point $p$.

Also, to ask whether it is bijective: It is clearly injective since the definition specifies one and only one value in the co-domain $(1,\infty)$ for each value in the domain $(0,1)$. To see if it is surjective, ask yourself: for which point $p$ in $(1,\infty)$ is there no point $x$ in $(0,1)$ such that $1/x = p$? If you can prove there is no such point, then you have shown it is surjective as well.

Finally, I believe any bijection which is continuous has a continuous inverse.

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    $\begingroup$ Regarding your last remark, see this $\endgroup$ – Giuseppe Negro Nov 20 '14 at 15:36
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Hint: Write down the definition of continuous. Rely heavily on the fact that $x\neq 0$ (in fact, if you close the interval around $1$ on both sides, it remains true).

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