Is Bounded subspaces of R be homeomorphic to R? Choose the correct statement(s):
(A) There is a continuous surjective function from [0, 1) to R;
(B) R and [0, 1) are homeomorphic to each other;
(C) There is a bijective function from [0, 1) to R;
(D) Bounded subspaces of R cannot be homeomorphic to R.
my answer : option1 is true  as  i take $$f(x) = \frac{1}{1-x}\sin\frac{1}{1-x}$$...
option2).  R and (0,1)  are homeormorphic so it   is false
option3) option 3 is false because R and [0, 1) are not  homeomorphic to each other so there is no bijection exist.
option 4) is True as subspace  $[0,1]∩R =[0,1]$ as $[0,1] $ is not homeomorphic to
 R.
is my answer is correct or not ...pliz verified and tell me the solution i will be glad and i woul be more thankful  .....
Thanks in advance
 A: (A) Your example is correct.
(B) The statement is false, but your argument isn't valid. Suppose there exists a continuous bijection $f:[0,1)\to\mathbb R$. Fix the point $f(0)\in\mathbb R$. Now you can deduce that $f([0,1))\subset [f(0),\infty)$ or $f([0,1))\subset(-\infty,f(0)]$ (if not, use IVT to get a contadiction. Then, $f$ is not bijective.
(C) There is a bijection from $[0,1)$ to $\mathbb R$. But it is a bit tricky to define. There is a question somewhere here on stackexchange, but I don't find it. The idea is the following:
For $n\in\mathbb N$ define $f\left(1-\frac1n\right)=n$. Next, you define $f$ bijective $\left(0,\frac12\right)$ into $(-\infty,1)$ and $\left(\frac{k}{k+1},\frac{k+1}{k+2}\right)$ into $(k,k+1)$ for all $k\in\mathbb N$.
(D) $\mathbb R$ as a $\mathbb R$-vectorspace is $1$-dimensional and has just the subspaces $\{0\}$ and $\mathbb R$. $\{0\}$ is not homeomorphic to $\mathbb R$ since its finite while $\mathbb R$ isn't. And $\mathbb R$ is not bounded.
If you are talking about subspaces in topological sense, then an example isn't enough to prove the statement that no bounded subspace of $\mathbb R$ is homeomorphic to $\mathbb R$. But a counterexample proves that the statement is wrong. And in topological sense you can consider the subspace $(0,1)\subset \mathbb R$, which is homeomorphic to $\mathbb R$.
