If F is Continuous and Bijection is F Inverse Continous

Show that $f, f^{-1}$ are continuous

Solution 1

A conceptual proof follows from the material in $\S$ 6.3 of my honors calculus notes:

Step 1: We are given that $f$ is bijective and increasing. So $f^{-1}$ exists and is moreover increasing: suppose not; then there are $y_1 < y_2 \in B$ with $f^{-1}(y_1) \geq f^{-1}(y_2)$. Then applying $f$ we get $y_1 \geq y_2$, contradiction. Thus the situation is perfectly symmetrical with respect to $f$ and $f^{-1}$, so it suffices to show that $f$ is continuous.

Step 2: We use the fact that every monotone function defined on $A$ has a left hand limit at every $c \in A$ -- namely $\sup \{ f(x) \mid x < c\}$ and a right hand limit -- namely $\inf \{ f(x) \mid x > c\}$, and the value $f(x)$ lies in between. (This is part of the Monotone Jump Theorem in $\S$ 6.3 of my notes.) Thus the only way we can have a discontinuity is if $\lim_{x \rightarrow c^-} f(x) < f(c)$ or $f(c) < \lim_{x \rightarrow c^+} f(c)$. But if either of these occurs, then $f(c)$ is not an interior point of $f(A)$, contradicting the hypothesis that $f(A)$ is open.

This answers the OP's question. I claim that it also proves that the inverse of a continuous function is continuous, at least in the case that the domain of $f$ is an interval. This is because every injective continuous function $f: I \rightarrow \mathbb{R}$ must be monotone: see $\S$ 5.6.3 of the notes. (There is a bit of combinatorial trickiness here.) Using the fact that every open subset of $\mathbb{R}$ is a disjoint union of open intervals -- which is not in the notes (I don't do any explicit topology whatsoever there) but is well known and not hard to show -- and that for every open interval $I$ and continuous $f$, $f(I)$ is an interval ($\S$ 6.2 of the notes) one sees that this extends to continuous functions on any open subset of $\mathbb{R}$, but this seems to be the longer way around this particular question.

Added: It is certainly not the case that any continuous bijection $f: X \rightarrow Y$ of topological spaces must have a continuous inverse. To get a counterexample, let $Y$ be your favorite non-discrete topological space, let $X$ be the same set endowed with the discrete topology, and let $f$ be the identity map. From this perspective the "automatic continuity" property of the inverse for continuous bijections on open subsets of $\mathbb{R}$ is surprising. It can be generalized to open subsets of $\mathbb{R}^n$ and then becomes a quite famous (and rather deep) theorem, Brouwer's Invariance of Domain. This can be generalized to topological manifolds. There are other "Open Mapping Theorems" in mathematics -- famously in complex analysis and Banach space theory -- but such results are highly prized, as they are the exception rather than the rule.

Solution 2

I think assuming that $y_n$ goes to y is assuming f is continuous.

The hypothesis is that f is strictly monotonic increasing. Let $x_n$ < x be a sequence which is monotonically increasing and converging to x. Then $f(x_n) \le f(x)$ for all n. Because {$x_n$} is monotone, the sequence {f($x_n$)} is as well and must have a limit, say y $\le$ f(x). However, y cannot be strictly less than f(x). If it is then because B $\subset$ R is open there must be a neighborhood of y in B containing a value w such that y < w < f(x).

Consider c = $f^{-1}(w)$. Because f is monotonic we have $x_n$ < c < x for all n. But by definition $x_n \rightarrow x$. So there is no such c, thus no such w and lim $f(x_n) = f(x)$.

Of course the same argument can be made using any decreasing sequence approaching x. Finally, if there is a sequence $a_n \rightarrow$ x which is not monotonic, you can repeat this argument with the lim sup and lim inf of $a_n$. So the sequence f($a_n$) must have a limit, and it must be f(x).

The conclusion is that f is indeed continuous.

If f is continuous so is $f^{-1}$, which follows from the definition of the inverse. Since Pete Clark is concerned about this statement, here is another approach.

Because f is monotone, $f^{-1}$ must be also. By definition $f^{-1}(B) = A$ and we are given that A is open. So the argument above can be applied to $f^{-1}$, making it continuous.

Comments

Let $A,B \subset \mathbb{R}$ be open, and $f:A\rightarrow B$ be surjective and strictly monotonic increasing. Show that $f,f^{-1}$ are continuous.

Proof: I first show $f$ is injective. Let $x,y \in A, \mbox{and } x\neq y.$ This means either $x<y \mbox{ or } y<x.$ As $f$ is monotonic increasing, $f(x)<f(y) \mbox{ or }f(y)<f(x).\Rightarrow f(x)\neq f(y)\Rightarrow f$ is injective. This shows $f$ is bijective.

To show $f$ is continuous, let $D \subset B$ be open. I need to show $f^{-1}(D)$ is open in $A$. Suppose not, i.e., $(f^{-1}(D))^{c}$ is not closed. $\quad\Rightarrow \exists$ a sequence $(x_n)_{n\in\mathbb{N}}$ in $(f^{-1}(D))^{c}$ that converges to $x$ which is in $f^{-1}(D)$. As $f$ is bijective, $\exists$ a unique $y_n,y$ for each $x_n$ such that $f(x_n)=y_n,f(x)=y,\forall n\in\mathbb{N}$, where $(y_n)_{n\in\mathbb{N}}$ is in $D^{c}, y\in D$. Here, $y_n\rightarrow y$. If it does not, this means $f^{-1}(y_n)=x_n$ does not converge to $f^{-1}(y)=x$, contradiction. $\quad \Rightarrow D^c$ is not closed. $\Rightarrow D$ is not open. Contradiction.

$f^{-1}$ can also be proved to be continuous in the same way as above.

I somehow get a feeling that I am not allowed to argue $y_n \rightarrow y$ because a priori, my argument, which I think, assumes $f$ is continuous, which I have not proven yet!

Is my proof correct or my doubt? Please help me get out of this situation!
- Yes,you are right. You assume that $f$ is continuous.To prove it, first show that $f$ is open ($\iff f^{-1}$ is continuous) using the fact that $f$ is strctly increasing,then show that $f$ is continuous.
- I would like to know from the downvoter his/her reason for voting down.
The result is false if $B$ is not open: see e.g. en.wikipedia.org/wiki/Invariance_of_domain (the one-dimensional case is much easier). So your argument needs to make reference to the openness of $B$. I couldn't see where you did that; could you please clarify?
Hi Pete, as usual you are most informative. In this case both A and B are in R. Is there an example where B $\subset$ R but $f^{-1}$ is not continuous? I did take a shortcut here in claiming it follows from f continuous; could have proved its continuity the same way as for f. I did look through the material you linked, but will have to spend some time to digest it.
Question: given that f and $f^{-1}$ are monotone (or there wouldn't even be an inverse); and given that the proof for continuity of f is the same as for the continuity of $f^{-1}$ why is it tricky to prove that $f^{-1} is continuous. Maybe in a more complicated domain?
If $f$ is continuous, then it is a true (nontrivial) fact that $f^{-1}$ is necessarily continuous: this is described in my answer. But we are not given that $f$ is continuous: we have to show that. (As mentioned in my answer, the hypotheses are symmetric on $f$ and $f^{-1}$, so there is no need to focus separately on the inverse function.) For this we certainly need $B$ to be open: just take any function having a jump discontinuity at (at least one) point.
@Betty: First, in order for a continuous bijection $f: A \rightarrow \mathbb{R}$ to be monotone, we need $A$ to be an interval. Let's assume that $A$ is an open interval: to get from this to the general case is not hard. Then it is true that a continuous bijection is monotone but that requires proof (real proof: one needs the completeness of $\mathbb{R}$). This is done in my notes as well; I find that argument a little trickier than the rest of the argument which is presented in my answer.
In contrast, given the monotonicity of $f$, the proof of continuity of the inverse is not bad: as I mentioned, it occurs at the end of $\S$ 6.3 of my notes. But you should compare with the earlier, more direct proof of the continuity of $f^{-1}$ given in $\S$ 5.6.4 of my notes, which is taken directly from Spivak's book. I find this proof tricky and unenlightening. Others may feel differently of course. Did you have in mind a proof of continuity of $f^{-1}$ without explicitly assuming that $f$ is monotone? What is it?
To be sure, I don't follow your argument for the proof of the continuity of $f$. As I mentioned, there are plenty of functions which satisfy all the hypotheses except the one that requires $B = f(A)$ to be open, and no such function can be continuous. So to get a correct proof you must use the openness of $B$. Did you use it? I couldn't see where.
My comments at the end about continuity of the inverse are just to illustrate that it does not follow trivially from the definition of continuous functions and inverses that the inverse of a continuous function must be continuous. I thought that you might have been suggesting that in your answer, but I wasn't (and still am not) sure.
You are absolutely right. It took me awhile to properly understand what you are saying. I am going to delete the answer until/unless I can fix it.
@PeteL.Clark -- see if you like this better.
If f is invertible it must be 1-1, and in R that means monotonic. So yes, I was assuming exactly that. Obviously monotonic will not work if you don't have an ordering on A and B. I understood this as the reason why continuous is defined as mapping open sets into open sets. However, the question here is in $R^1$ and that is what I answered.
@Betty: The functions are defined on open subsets of $\mathbb{R}$, not necessarily on all of $\mathbb{R}$ or on a single interval. An open subset of $\mathbb{R}$ is a disjoint union of intervals, and a bijective function needs to be monotonic when restricted to each interval: but it could for instance be increasing on one interval and decreasing on another. (Of course no one is talking about monotonic functions on nonordered sets!)
The above is mostly a technicality; more significantly, the point is that showing that a bijective continuous function from an interval to an interval requires the completeness of $\mathbb{R}$: e.g. if you restricted to intervals in $\mathbb{Q}$ the conclusion need not hold. This is a sign that the proof is not trivial: it uses the Intermediate Value Theorem but is still a bit tricky. I gave a careful treatment in my note. Again, maybe you have a different, simpler proof in mind: I would be happy to see it. But if you want to claim it's obvious: no, I disagree.
Finally, a continuous function is not defined as mapping open sets into open sets. That's the definition of an open mapping. Clearly a continuous function need not be an open mapping: take a constant function. More subtly, there are -- in contexts more general than subsets of Euclidean space -- continuous bijections which are not open mappings.
It is much improved. I didn't understand the part of your argument which deals with an arbitrary sequence converging to $x$: if such a sequence contains some terms smaller and some terms larger than $x$, then the argument does not seem to go through. However this can be fixed in several ways: for instance it is not hard to show that for any function $f: A \rightarrow \mathbb{R}$ and $c \in A$, $\lim_{x \rightarrow c^-} f(x) = f(c)$ if and only if for every increasing sequence $x_1 < \ldots < x_n < \ldots c$, $f(x_n) \rightarrow f(c)$; and the analogous statement for right hand limits.
And of course the argument about the symmetry between $f$ and $f^{-1}$ is the one given in my answer, so I find it unobjectionable. With these changes your becomes essentially the same as mind, only with continuity always being checked using sequences. I certainly find such an answer to be correct! I am a little disappointed that you still seem to think that "$f$ continuous implies $f^{-1}$ continuous" is somehow trivial (and that $f$ continuous bijective implies $f$ monotone is trivial). It might be good to enlist another opinion on this.
of course the open set A could consist of numerous intervals, and of course f could be monotonic in different directions on each interval. So I wasn't being clear and specific enough. I see that the completeness of R is essential. You are right, a constant function is not an open mapping. Can you link me to ( or suggest) an example of a continuous bijection which is not an open mapping?
@Betty: My answer above gives an example of a continuous bijection between topological spaces which is not open. There are no such examples when the spaces are topological manifolds...but that's quite a deep theorem in algebraic topology.
As I said above, it is entirely possible that my argument that a bijective continuous function between intervals in $\mathbb{R}$ is monotone is too involved. This is not such a standard result in textbooks, so I'm not sure what other people do. I googled for it and found this proof, which is very close to the one in my notes: proofwiki.org/wiki/…. I don't think there's anything deep or even especially interesting going on here, but I know from experience that many undergraduates would have trouble nailing down the details.
Have a good holiday.

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If F is Continuous and Bijection is F Inverse Continous

Show that $f, f^{-1}$ are continuous

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Solution 2

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