In the middle of talking about primitives and the Cauchy integral theorem, my Complex Analysis teacher came up with this sentence:
This reasoning can be done in any simply connected set, because in $\mathbb{R}^2$ and $\mathbb{R}^3$ every simply connected set is path connected, this is a true theorem.
Now, I always conceived simply connectedness as having path connectedness in its definition, so I wondered what he meant. I looked "simply connected" up on Wikipedia and I found confirmation. My idea is that he thinks simply connected simply means:
A space is simply connected if every closed path can be contracted to a point.
That is, he removes path connectedness from this definition. Regardless of whether this removal is correct or not, I was wondering:
Is it true that in $\mathbb{R}^2$ and $\mathbb{R}^3$ the above definition implies path connectedness? Is there any property of a topological space that grants the implication? If so, which, and how do you prove it grants the implication?
PS The reasoning was that if we have a holomorphic function in a simply connected domain, then it has a primitive because we just fix a point in the domain and integrate the function along any path, or more precisely if $f:\Omega\to\mathbb{C}$ is defined on a simply connected $\Omega$ it has a primitive in $\Omega$ since we just define $F(z)$ to be the integral of $f$ from any fixed point $z_0$ to $z$, along any path. That is because all paths with the same endpoints are homotopic due to the contractibility of their junction because of simple connectedness, so the integral in question does not depend on the particular path chosen. This is for the sake of completeness, as a sort of background. Not part of the question.
Posting this to sum up the comments and take this question off the unanswered list.
In the following, I will refer to the property that all continuous closed paths are homotopic to a point as "the property".
In general, the property doesn't imply path connectedness, even for open subsets: as iwriteonbananas points out, just take the union of two disjoint open balls, those have the property but are not path connected.
One way to get around this is to suppose my professor was considering open connected subsets of $\mathbb{C}$, as Potato suggested. In that case, since $\mathbb{C}$ is locally path connected (i.e. has a basis of p.c. open sets -- the balls), then an open connected set of $\mathbb{C}$ is path connected. So «the real theorem», as my professor put it, is that $\mathbb{C}$ (or $\mathbb{R}^2$) is l.p.c., which is the key point in getting that open connected subsets are p.c., or that any open set with the property is simply connected.
Examples of connected but not p.c. spaces that came up in the final comments were the Topologist's sine and the flea and comb (which I always knew with that name, but is apparently aka "deleted comb").
As an extra, xhmi pointed out in his comment that contractibility implies p.c.-ness. Indeed, if $X$ is contractible, i.e. there exists $F:[0,1]\times X\to X$ a continuous function such that $F(0,x)=x$ for all $x$ and $F(1,x)=y$ for all $x$ for some $y\in X$, then take any $x\in X$ and $\gamma(t)=F(t,x)$ is a continuous path from $y$ to $x$, meaning every point is connected via a continuous path to $y$, and by joining paths any point is connected to any point, which is precisely path-connectedness.