Continuous bijection from $(0,1)$ to $[0,1]$ Does there exist a continuous bijection from $(0,1)$ to $[0,1]$? Of course the map should not be a proper map.
 A: Let $f:(0,1) \rightarrow [0,1]$ be continuous and surjective. (Actually, we just need to suppose that $0$ and $1$ are in the image of $f$.) Let $a,b \in (0,1)$ such that $f(a)=0$ and $f(b)=1$.  Let $I=[a,b]$ if $a<b$ or $I=[b,a]$ if $b<a$. Then, by the intermediate value theorem, $f(I)$ is an interval that contains $0$ and $1$ and so $f(I)$ contains $[0,1]$, which implies $f(I)=[0,1]$. But then $f$ cannot be injective because $(0,1)\setminus I$ is nonempty.
A: Since Theo gave an answer I am going to be nitpicking and add one remark. When speaking about continuity (especially when tagging under [topology]) it is best to mention the topology you are working with. In this case, you mean in the standard topology.
Otherwise, consider the discrete topology, i.e. every set is open:
Let $f\colon [0,1]\to (0,1)$ be any bijection, it is continuous since all sets are open, the preimage of an open set is an open set, thus $f$ is continuous.
A: Recall the following result:
Proposition 1: Let $f: (a,b) \to (c,d)$ be an injective continuous mapping from one open interval to another. Then the image of $f$ is an open interval.
Now assume we have a continuous bijection $f: (0,1) \to [0,1]$. The function $f$ maps some point in $(0,1)$ to $0$ and another point to $1$. If we remove these points we can restrict $f$ and define an injective and surjective continuous mapping
$\tag 1 f: (0,a) \sqcup (a,b) \sqcup (b,1) \to (0,1)$
on three non-overlapping intervals.
By proposition 1 the image of each of these three intervals is an open interval of $(0,1)$. Since $f$ is also surjective, we have a partition of $(0,1)$ into three open subsets. But it is not possible to express a connected topological space in such a manner.
A: No. If $f:(0,1) \to [0,1]$ were continuous and bijective, there would be a unique point $x \in (0,1)$ such that $f(x) = 1$. However, since $f$ is continuous, the intervals $[x - \varepsilon, x]$ and $[x, x + \varepsilon]$ would be mapped to intervals $[a,1]$ and $[b,1]$, say. By bijectivity we'd have $a, b \lt 1$. Thus every value strictly between $\max{\{a,b\}}$ and $1$ would be assumed at least twice, contradicting bijectivity.
A: Suppose that $f:(0,1) \rightarrow [0,1]$ is 1-1 and continuous.  By the intermediate value theorem, the image of any interval under $f$ is an interval.  Since $f$ is 1-1, it is either (strictly) monotone increasing or decreasing.  Hence, $f(0,1)$ is an interval.  Without loss of generality, assume $f$ is increasing; were it not this analysis would apply to $1 - f$.
Suppose now that $f$ is onto; then we must have some $t\in(0,1)$ with $f(t) = 1$.  Because $f$ is strictly monotone increasing, we would have to have $f(s) > 1$, for $t \le s < 1$.  This violates the premise that $f(0,1) 
\subseteq [0,1]$.  Hence, $f$ cannot be onto.
A: There does not exist  a continuous bijection from (0,1) to [0,1]. Indeed, let $f$ be such a function. Let consider a sequence  $x_n=1-1/n$. Then from the sequence $(f(x_n))$ we can choose a subsequence $(f(x_{n_k}))$  which is convergent. Let denote this limit by $y$. Obviously, $y \in [0,1]$. Since $f^{-1}$ also is continuous, we get $f^{-1}(y)=\lim_{k \to +\infty}f^{-1}(f(x_{n_k}))=\lim_{k \to \infty}x_{n_k}=1$. But $1 \notin (0,1)$.  
Remark(Why $f^{-1}$ must be continuous under our assumption?)  By our assumption $f:(0,1)\to [0,1]$ is continuous bijection. Then  $f:(0,1)\to [0,1]$ must be injective and continuous which following invariance of domain (see, http://en.wikipedia.org/wiki/Invariance_of_domain   is  homeomorphism. Hence $f^{-1}: [0,1]\to (0,1)$ is continuous.
