Alternative definition of simple connectedness Today in (what can typically be called a Calc III) class, our lecturer gave us the following definition :

Lecturers Definition of Simple Connectedness : A simple connected region in the plane is a connected region $D$ such that every simple closed curve in $D$ encloses only points in $D$

This isn't really a rigorous definition, what I'm assuming the lecturer means is that the plane is $\mathbb{R}^2$ with the usual metric topology. An $D \subseteq \mathbb{R}^2$ is connected.
But the only definition I know of simple connectedness is the following. A topological space $X$ is simply connected if it is path-connected and $\pi_1(X, p) = \{[c_p]\}$. (i.e. the fundamental group of $X$ is tribial)
So is the usual definition of simple-connectedness above equivalent to my lecturer's definition? I don't think so.
Take $\mathbb{S}^1$ as an example, $\mathbb{S}^1$ is path-connected and hence connected and it is very well known theorem that $\pi_1(\mathbb{S}^1) = \mathbb{Z}$ and hence $\mathbb{S}^1$ is not simply connected.
But by the definition my lecturer has given quoted above, (if encloses means contains) if we take $D = \mathbb{S}^1$, then "every simple closed curve in $D$" corresponds to paths around the perimeter of the circle with the same starting and ending points, and hence only contains points in $D$ and is simply connected.
So is my lecturer's definition wrong? I think so.
 A: Your instructor's definition qualifies as "morally wrong but technically correct". More precisely: It is morally wrong since this is not at all how one defines simple connectivity in general (even for non-planar surfaces). It is however, useful to teaching purposes, since "students understand it much better this way". (Which is what an elementary school teacher said when asked why was she teaching students to add numerator to numerator and denominator to denominator while explaining the addition of fractions.) Nevertheless, this definition is equivalent to the standard one, but proving this goes well beyond a Calc-III class. Here is a sketch of one possible proof:
Suppose that $S$ is a connected planar surface such that every Jordan curve $J\subset S$ bounds a disk contained in $S$. I will think of $S^2$ as the 1-point compactification of $R^2$, $S^2=R^2 \cup \{\infty\}$ (use the stereographic projection.)  Let us prove that 
$E=S^2 -  S$ is connected. Suppose not. Take a partition of $E$ in two disjoint nonempty closed subsets $E= A\sqcup B$. Let $\chi: E\to \{0,1\}$ be the characteristic function of $A$. Now, use the Tietze-Urysohn extension theorem (which you may learn in an undergraduate general topology class) to extend $\chi$ to a continuous function $f: S^2\to {\mathbb R}$. Modify the extension (using the smooth approximation theorem) to make it smooth in $S$. (I still call the new function $f$.) Use Sard's regular value theorem to find $t\in (0,1)$, a regular value of $f$. Then $f^{-1}(t)$ is a compact 1-dimensional submanifold $M\subset S$. Each component $J$ of $M$ is a Jordan curve. At least one of these curves, $J$, will separate $A$ from $B$, just by the intermediate value theorem (Calc-I - finally I can quote something that you had seen before). Therefore, $J$ cannot bound a disk in $S$. A contradiction. Thus, $E=S^2 -S$ is connected. Now we use some algebraic topology (which you are very unlikely to see as an undergraduate). (a) Use Alexander duality to conclude that $H_1(S)=0$. (b) Prove that $\pi_1(S)$ is free (as for any noncompact connected surface: see here). (c) Use the Hurewicz theorem to conclude that $\pi_1(S)$ is trivial since $H_1(S)$ is isomorphic to the abelianization of $\pi_1(S)$. 
Now, you are done proving that your instructor's definition implies that $S$ is simply connected in the usual sense. For the converse, you can use, for instance, the Riemann mapping theorem (which you might see in an undergraduate complex analysis class: if your class is called Analysis-III rather than Calc-III, then you might see the RMT quite soon, at least the statement), to conclude that every simply connected planar surface $S$ is  homeomorphic to $R^2$ (as it is either equal to ${\mathbb C}$ or is conformal to the unit disk). Lastly, use Jordan curve theorem (which one again can prove via Alexander duality) to conclude that every Jordan curve bounds a disk in $S$. 
Just for fun, you may want to ask your instructor if this is the proof that he/she had in mind. 
A: The points "enclosed" by a simple closed curve in $\mathbb{R}^2$ are the bounded component of its complement in $\mathbb{R}^2$.  That is, if $C\subset\mathbb{R}^2$ is (the image of) a simple closed curve, $\mathbb{R}^2\setminus C$ always has two connected components (by the Jordan curve theorem), one of which is bounded.  The points "enclosed" by $C$ are the points of the bounded component.  So for instance, if $C=\mathbb{S}^1$, it encloses the entire open unit disk.  This means that according to youre lecturer's definition, a simply connected subset of $\mathbb{R}^2$ which contains $\mathbb{S}^1$ must also contain the entire unit disk.
Your lecturer's definition of "simply connected" is in fact equivalent to the standard one for connected open subsets of $\mathbb{R}^2$.  The proof is difficult.  As a sketch of a proof, any simple closed curve in $\mathbb{R}^2$ cannot be contracted to a point in the complement of any point it encloses, and so if $D$ is simply connected by the usual definition it must satisfy your lecturer's condition.  Conversely, if $D$ is not simply connected, any nontrivial loop in $D$ can be homotoped to one with only finitely many self-intersections, and thus can be broken into finitely many simple closed curves.  If all enclosed points of all these simple closed curves were in $D$ then the loop would be nullhomotopic in $D$, so $D$ cannot satisfy your lecturer's definition.
