Intermediate Value Property I am trying show that the function $f:[0,1]\to \mathbb{R}$ defined by $f(x)=\sin \dfrac{1}{x}$ if $x\neq 0$ and $f(0)=0$ possesses IVP. Though it looks easy, but I am not getting any clue how to start with. Any help would be appreciated.
 A: Show that there exists a subset $A$ of $(0,1]$ such that $f(A)=f([0,1])$ and such that $f|_A$ ($f$ restricted to the domain $A$) is a continuous function. You may then apply the intermediate value theorem to $f|_A$.
Note that the above proves that $f$ has the property that you mentioned in the comments, but this is not what one would usually call the IVP. The usual intermediate value property is that for any two values $a$ and $b$ in the domain of $f$, and any $y$ between $f(a)$ and $f(b)$, there is some $c$ between $a$ and $b$ with $f(c) = y$. We call functions which satisfy the IVP Darboux functions. This question highlights the fact that the set of Darboux functions $[0,1]\rightarrow\mathbb{R}$ is a proper superset of the set of continuous functions $[0,1]\rightarrow \mathbb{R}$.
Assuming $a<b$, the above proves the IVP for $a=0$, $b=1$. For $b<1$, you will need to show that an $A$ exists as above, but such that $A\subset (0,b)$. It also remains to show that the IVP holds for $a\neq 0$ but this case is handled rather easily by simply restricting $f$ to the interval $[a,b]$ and noting that $f|_{[a,b]}$ is continuous. 
A: Consider the interval
$$I_{k} = \left[ \frac{1}{\left(2k+\frac{3}{2}\right)}, \frac{1}{\left(2k+\frac{1}{2}\right) \pi}\right]$$
notice that $-1\leq f(I_{k}) \leq 1$ for all $k\in \mathbb{N}$ and that $I_{1}\subseteq I_{2}\subseteq \dots$
A: You want to show -- despite $f$ is not continuos -- that every value between $f(0)=0$ and $f(1)=\sin(1)$ is attached by $f$.  That's simple: consider the restriction of $f$ on $[1/\pi,1]$.  Can you work from here applying IVT on that restriction?
