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I was looking for examples of bijective continuous functions infollowing cases

  1. Does there exist Bijective continuous function $(0,1) \rightarrow \mathbb{R}$?

Yes, $f(x)=\frac{2x-1}{x-x^2}$

  1. Does there exist Bijective continuous function from $\mathbb{R} \rightarrow (0,1)$ ?

Yes, as $(0,1) and \mathbb{R}$ are homeomorphic so inverse of $f(x)$ defined above will be example such a function.

  1. Does there exist a bijective continuous function from $[a,b] \rightarrow \mathbb{R}$?

I think No, as continuous image of compact set is not compact.

  1. Does there exist a bijective continuous function from $\mathbb{R} \rightarrow [a,b])$ ?

For this clearly it can be seen

a) continuous image of connected set is connected.

b) Inverse image of closed set under continuous map is closed.

So there exist such a function but i can not construct it.

  1. Does Does there exist a bijective continuous function from $ (a,b) \rightarrow [a,b])$?

No. as inverse image of closed set is not closed so there can not exist any continuous bijective function.

But as $(a,b)$ is homeomorphic to $\mathbb{R}$So if we construct a function $\mathbb{R \rightarrow} [a,b]$ will that function work as an example of a function $ (a,b) \rightarrow [a,b])$?

  1. Does Does there exist a bijective continuous function from $ [a,b] \rightarrow (a,b))$?

No. as inverse image of open set is not open so there can not exist any continuous bijective function.

Please correct me if I am wrong anywhere .And give examples of $(4)$ if any exists. Thanx very much for your time

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  • $\begingroup$ I've always thought of Probability distributions being from R to [0,1] $\endgroup$ – shai horowitz Jun 28 '16 at 5:20
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  1. No. Suppose there exists such a function $f$. Say $f(x_0) = a$, then $f(x) > a$ for both $x<x_0$ and $x> x_0$, in contradiction with $f$ being a bijection.

5 and 6. Your arguments are invalid. Given continuous function $f:X\to Y$, the pre-image of an open(closed) set in $Y$ under $f$ is open(closed) in $X$. Here $(a,b)$ is indeed closed in $(a,b)$, and $[a,b]$ likewise. That no continuous function maps $[a,b]$ onto $(a,b)$ can be easily seen by noticing that $[a,b]$ is compact while $(a,b)$ isn't. That no continuous bijection maps $(a,b)$ to $[a,b]$ can be reasoned similarly as in 4.

Also be aware that in general a continuous bijection doesn't have to be a homeomorphism, so you probably want to refine your arguments in 2. That said, we do have invariance of domain theorem in Euclidean spaces.

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  • $\begingroup$ Thanx. I get your reasoning for 4. 5 and 6. But what if left the condition of continuity then can we find a bijective function in all cases? It seems yes to me $\endgroup$ – slow but keen learner Jun 28 '16 at 5:13
  • $\begingroup$ Yes. See math.stackexchange.com/questions/28568/…, for example. There should be many answers on this site. $\endgroup$ – Qiyu Wen Jun 28 '16 at 5:15
  • $\begingroup$ ok I was unaware of Invariance of domain theorem. Now with the help of this i can directly say $f^{-1}$ is continuous as $(0,1) \subset \mathbb{R}$. Now if we left the condition of bijection then identity function will be example of (3) and can we find example for (4 ) $\endgroup$ – slow but keen learner Jun 28 '16 at 5:21
  • $\begingroup$ Constant functions. Or if you want surjection, let $f$ be identity on $[a,b]$, $f(x) = a$ for $x<a$, $f(x) = b$ for $x>b$. $\endgroup$ – Qiyu Wen Jun 28 '16 at 5:25
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Your proposed $f$ for (1) is not a bijection. One that is:

$f(x)=\tan(\pi x-\frac{\pi}{2})$

Crucially, any such $f$ must be monotone (in-fact it cannot have derivative 0 over any interval, only at single points) (why?)

For (2), just invert (the corrected) function for #1.

(3) is no because $[a,b]$ is compact, but $\mathbb{R}$ is not (cannot satisfy surjectivity and continuity at the same time for reason you gave).

(4) is also no, but the reasoning is slightly different (see @QiyuWen's answer).

(5) You're reasoning is wrong, it's for the same reason as in (4) (this has to do w/ monotonicity again).

(6) Appeal to compactness again ($(a,b)$ is not compact; why?).

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  • $\begingroup$ As an additional comment: $(a,b)$ is open but not closed as a subset of $\mathbb{R}$, but as a subset of itself (under the induced subspace topology), $(a,b)$ is both open and closed (the entire space is always both open and closed, same for the empty set). This is an important subtlety which comes from the definition of subspace topology: The subspace topology on any $S\in X$, $X$ a topological space, is formed by taking the intersection of each open set of $X$ with $S$ to form the open sets of $S$. $\endgroup$ – Justin Benfield Jun 28 '16 at 6:57

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