Topology: Open Sets, Domains, Simply Connected
Kaplan §2.2 - the vocabulary the rest of vector calculus assumes.
That distinction - which loops can contract, which can't - is the difference between a region where every conservative-looking field really does come from a potential, and a region where it might not. Before we can even ask the question precisely, we need a small vocabulary: neighborhood, open, closed, bounded, domain, simply connected. Kaplan defines all of these in §2.2, and they recur on almost every page after.
Neighborhoods - the basic move
Topology starts with one simple idea: zoom in on a point and ask what's nearby. A neighborhood of a point $p = (x_1, y_1)$ is the set of points within some chosen distance $\delta > 0$ from $p$:
Geometrically, that's the open disk of radius $\delta$ centered at $p$ - the rim itself is not included (the inequality is strict). In three or more dimensions, replace the disk by a ball; the definition reads the same with more coordinates. Kaplan calls this $\delta$ the radius of the neighborhood, and the only thing we ever do with it is make it small.
That is the entire primitive. From here on, every topological idea on this page is phrased as: "for every point in the set, some neighborhood does (or does not) do such-and-such."
Open sets
Suppose we have a set $E$ in the plane. Pick any point $p$ in $E$. Can we find a neighborhood of $p$ that lies entirely inside $E$? If yes for every point of $E$, the set is open.
The intuition: every point has wiggle room. You can perturb it a little in any direction and stay in $E$.
Examples of open sets: the open disk $x^2 + y^2 < 1$, the open square $|x| < 1, |y| < 1$, the right half-plane $x > 0$, the entire plane. The strict inequality is what lets us shrink $\delta$ enough to stay inside.
Non-examples: the closed disk $x^2 + y^2 \le 1$ is not open - take $q = (1, 0)$ on the rim, and any neighborhood of $q$ contains points with $x > 1$, which are outside. The same trick rules out any set that swallows a piece of its boundary.
Closed and bounded
Two more pieces of vocabulary, then the deeper notions.
A set $E$ is closed if its complement (everything in the plane that is not in $E$) is open. Equivalently - and more usefully - $E$ is closed if it contains all of its boundary points. A point $q$ is a boundary point of $E$ if every neighborhood of $q$ touches both $E$ and its complement: it sits on the edge.
So the closed disk $x^2 + y^2 \le 1$ is closed (it contains its rim). The open disk $x^2 + y^2 < 1$ is not closed (the rim is missing). The half-open set $0 < x \le 1$ is neither open (it has the rim $x=1$ stuck on) nor closed (it's missing the rim $x=0$). "Not open" does not mean "closed" - most sets are neither.
A set is bounded if it fits inside some neighborhood - some single ball of finite radius contains all of it. The unit square is bounded; the right half-plane $x > 0$ is open but not bounded; the punctured plane minus the origin is open and unbounded.
A set that is both closed and bounded is what later chapters call compact. The closed disk is compact; the open disk is not. Continuous functions on compact sets attain their max and min - that's the punchline we'll cash in repeatedly.
Domain - Kaplan's specific sense
Here is where the language gets tricky. In Kaplan, a domain is a non-empty open set with one extra property: any two points in it can be joined by a polygonal path - a finite chain of line segments - that stays inside the set.
The "broken line" condition is just connectedness in disguise: a domain is one piece, not two disjoint open blobs. The set $\{x \neq 0\}$ in the plane is open, but it splits into the left and right half-planes - you cannot get from $(-1, 0)$ to $(1, 0)$ without leaving the set, so it's not a domain in Kaplan's sense.
A region is a domain plus, possibly, some or all of its boundary. If we throw in all the boundary points, we get a closed region. So the open disk is a domain; the closed disk is a closed region; the open disk together with half its rim is a region but neither open nor closed. Kaplan often writes a closed region as $R = D \cup \partial D$.
Most domains we'll meet are described by inequalities - $x^2 + y^2 < 1$, $0 < x < 1$ and $0 < y < 1$, $xy < 1$. The boundary is the corresponding equality, and the closed region is the non-strict inequality.
Simply connected
Connectedness says the region is one piece. Simple connectedness adds a second condition: there are no holes.
Picture the curve as a rubber band lying in $D$. If you can always pull it tight to a single point without snagging on something outside $D$, the region is simply connected. A snag is a hole.
Examples of simply connected planar domains: the open disk, the open square, any convex region, the whole plane, even a half-plane. Non-examples: the annulus $1 < x^2 + y^2 < 4$, the punctured disk $0 < x^2 + y^2 < 1$, the plane minus a finite set of points.
For planar regions there is a clean equivalent: $D$ is simply connected if and only if every simple closed curve $C$ in $D$ has its entire interior contained in $D$. In other words, no loop in $D$ encloses a forbidden region. That picture - "no holes" - is the one to keep in mind.
This is the missing hypothesis in Green's theorem and in the converse "curl-free implies conservative." A curl-free field on a simply connected domain has a potential. Drop the simply-connected assumption and the converse can fail - the famous example is $-y/(x^2+y^2)\,\mathbf{i} + x/(x^2+y^2)\,\mathbf{j}$ on the punctured plane.
A subtlety in three dimensions
In the plane, "simply connected" lines up with intuition: no holes. In space, the same word means the same thing - every closed curve shrinks to a point - but the geometry is more forgiving, and we have to think harder about what counts as a hole.
Take $\mathbb{R}^3$ minus a single point. Is it simply connected? Surprisingly, yes. Any loop, no matter how it threads near the missing point, can be slid off in the third dimension and contracted. There is no "hole" in 3D - a single missing point is a zero-dimensional defect, and a one-dimensional loop can step around it.
Now take $\mathbb{R}^3$ minus an entire line - say, the $z$-axis. A horizontal loop around the line still cannot be shrunk: any continuous deformation has to cross the $z$-axis somewhere, which is forbidden. So $\mathbb{R}^3$ minus a line is not simply connected. The line is a one-dimensional defect, and it traps two-dimensional loops the same way a missing point traps loops in the plane.
This distinction is exactly what separates two famous vector fields:
- The central inverse-square field $\mathbf{u} = \dfrac{\mathbf{r}}{|\mathbf{r}|^3}$ is defined on $\mathbb{R}^3 \setminus \{0\}$. That domain is simply connected. The field is curl-free, and (by the converse of Stokes' theorem) it has a potential globally - $\phi = -1/r$.
- The magnetic field around a long straight wire along the $z$-axis is essentially $\mathbf{u} = \dfrac{-y\,\mathbf{i} + x\,\mathbf{j}}{x^2 + y^2}$, defined on $\mathbb{R}^3$ minus the $z$-axis. That domain is not simply connected. The field is curl-free, but no global potential exists - circulation around the wire is non-zero. This is exactly Ampère's law in disguise.
The topology is doing real physical work. The vocabulary on this page is what lets us state cleanly when a "locally conservative" field is also globally conservative.
Practice Problems - §2.2
From Kaplan, problems after §2.4 (covering the topology of §2.2).
$\log$ requires a strictly positive argument, so we need $x^2 + y^2 - 1 > 0$, i.e. $x^2 + y^2 > 1$. The set is the exterior of the unit circle - everything strictly outside.
Yes. Every point $p$ with $|p| > 1$ has a small enough neighborhood that still satisfies $|q| > 1$ - shrink $\delta$ until $\delta < |p| - 1$. The strict inequality is what makes openness work.
No. The unit circle $x^2 + y^2 = 1$ consists of boundary points of our set, and none of them belong to it (the strict inequality excludes them). A closed set must contain its boundary.
No. The set extends to infinity in every direction; no single ball of finite radius can swallow it.
Yes. It is open (Step 2) and connected by broken lines: any two points outside the unit disk can be joined by going first radially outward to a large circle, then around, then back in - a finite polygonal path that never enters the disk. So this is a domain in Kaplan's sense. (It is not simply connected, though - a loop around the unit circle can't contract.)
The square root requires $1 - x^2 - y^2 \geq 0$, i.e. $x^2 + y^2 \leq 1$. The set is the closed unit disk.
No. The point $p = (1, 0)$ is in the set, but every neighborhood of $p$ contains points with $x^2 + y^2 > 1$, which are outside. Boundary points kill openness.
Yes. The complement is $x^2 + y^2 > 1$, which we just argued is open in problem 1. The complement of an open set is closed.
Yes. The whole set sits inside the neighborhood $N_2(0)$ - a disk of radius $2$ around the origin clearly contains the unit disk.
No - a domain has to be open. This is a closed region instead: the closed unit disk equals the open disk (a domain) plus its boundary circle.
The definition says: for every point $p$ in the set, some neighborhood lies inside. The empty set has no points, so the condition is vacuously satisfied - there's nothing to check. So $\emptyset$ is open.
The complement of $\emptyset$ is the whole plane $\mathbb{R}^2$. Pick any point $p$; the neighborhood $N_1(p)$ is contained in $\mathbb{R}^2$ trivially. So $\mathbb{R}^2$ is open, which means $\emptyset$ - its complement - is closed.
By the same argument run in the other direction: $\mathbb{R}^2$ is closed iff its complement $\emptyset$ is open, and we just showed it is. So $\mathbb{R}^2$ is both open and closed. Topologists call sets with this property clopen; in the plane the only clopen sets are $\emptyset$ and the whole plane, and that fact is essentially the connectedness of $\mathbb{R}^2$.