Is a simply closed curve in $R^2$ a bounded closed set? I am studying calculus now.I haven't learnt topology yet. It seems that the book defaults this conclusion with out proof.I am now confused about the definition of "curve".I have searched "curve" on Wiki and it says "A curve is a topological space which is locally homeomorphic to a line."
Is a simply closed curve in $R^2$ a bounded closed set ? Why? Who can explain this plainly?Thanks.
 A: At the calculus level the implicit definition of "simple closed curve" is probably something like "a continuous function $f$ from the unit interval to the plane such that $f(0)=f(1)$". The image of such a function is a bounded closed set.
A: For any simple closed curve $C \subset \mathbb{R}^2$, we can find a continuous parametrization $\beta: \mathbb{R} \rightarrow \mathbb{R}^2$ of it, where $0 \leq t \leq a$ for some $a \in \mathbb{R}$.  Given that the function $||\beta(t)||$ is also continuous, in particular on the domain $[0, a]$, apply the extreme value theorem to get that $C$ is bounded.

If you're getting into the more theoretical side of calculus, you can actually tackle both closedness and boundedness with one simple theorem (for which the extreme value theorem is a corollary):

The continuous image of a compact set is compact.

The compact subsets of $\mathbb{R}^n$ are precisely the sets which are closed and bounded$^\dagger$.  Because $[0,a]$ is such a set in $\mathbb{R}$, and because $\beta$ is continuous, then, per the theorem, $\beta\Big([0,a]\Big) = C$ is also compact and hence closed and bounded.

$^\dagger$This is the Heine-Borel theorem.
A: A (planar) curve is just a continuous $f:[0,1] \to \mathbb{R}^2$. The curve is closed iff $f(0) = f(1)$ (we "end" where we "start", if we see $[0,1]$ as a time parameter, and trace the image points as we go from $0$ to $1$). The curve is simple closed, iff $f(t) \neq f(t')$ whenever $t \neq t'$ and $\{t,t'\} \neq \{0,1\}$, in words: no point besides the start/end point is visited twice.
Your text actually means $f[[0,1]]$ (the image of $f$) as a subset of the plane, and this is then compact as the continuous image of a compact set $[0,1]$. The image of a simple closed curve is indeed just homeomorphic to the unit circle, and locally homeomorphic to the reals. 
In general though, there are space-filling curves, e.g. continuous maps from $[0,1]$ onto $[0,1]^2$, making the square a "curve" (or "curve-image"), which is quite non-intuitive. This inspired topologists into defining and studying topological dimension (as being "parametrised by one real number" is not the same as "one-dimensional", apparently, so a better definition is required).
