Connectedness is defined by the absence of something, so showing that a space is connected means ruling out every separation, while showing that it is disconnected means producing one. Producing one is the easy half, and this lesson is about doing it well: one set that is open and closed at once is enough, and the sets that do the splitting are often cuts at a point the space does not have.

To prove a wall has no door you must walk the whole wall. To prove it has one you point at a door. The two jobs are not the same size, and only one of them is finished by an example.

The space here is $\mathbb{Q}$, and the cut is made at a point that is not in it. The rationals below $\sqrt2$ and the rationals above it are both open in $\mathbb{Q}$, neither is empty, they share no point, and together they are everything — a separation. The gap the picture shows is invisible from inside the space, which is why the separation has to be written in open sets rather than pointed at.

A space is disconnected when it has a separation: open sets $U$ and $V$, neither empty, sharing no point, with $U\cup V=X$. Equivalently, it has a subset other than $\varnothing$ and $X$ that is both open and closed — a clopen subset. Either form is produced by exhibiting one example, and each form is easier to reach in different spaces.

• $\mathbb{Q}$ is disconnected: cut it at any irrational point. Both halves are open, because an irrational point has a whole interval of room round it inside neither half.
• $\mathbb{R}\setminus\{0\}$ is disconnected by the negatives and the positives, and $[0,1]\cup[2,3]$ by the two intervals, each of which is open in the subspace.
• Any discrete space with two or more points is disconnected: every subset is open, so every subset is clopen.
• The Sierpiński space $\{a,b\}$ with $\tau=\{\varnothing,\{a\},X\}$ is connected. Its only clopen sets are $\varnothing$ and $$: is open and not closed, closed and not open. So a space can be connected while being very far from a line.
• $\mathbb{Q}$ is totally disconnected: the only connected subsets it has are the empty set and single points, since any two rationals are cut apart by an irrational between them.

Cut the rationals at the square root of two, which is not itself rational. The rationals below it and the rationals above it are traces of two open rays, so both are open in the subspace. Neither is empty, since 1 lies in the first and 2 in the second. No rational lies in both, because no rational equals the cut point and a number is on one side or the other. And every rational lies in one of them, for the same reason: the cut point is missing from the space, so nothing is left over in the middle. That is a separation, and it shows the space is disconnected.

Cut the space at an irrational point and take the two sides.

Each is the trace of an open ray, and traces of open sets are the open sets of a subspace.

Both sides hold a rational, so neither is empty.

No rational equals the cut point, so each lies strictly on one side and on one side only.

The pair is a separation, and one separation is all that disconnectedness asks for.

To show a space is disconnected, produce one separation, or equivalently one clopen set that is neither empty nor everything.

• A separation is exhibited; connectedness has to be argued for every possible pair.
• $\mathbb{Q}$, a punctured line and any discrete space with two or more points are disconnected.
• A clopen subset other than the two extremes is the same thing as a separation.
• The Sierpiński space is connected, so connectedness does not need a line to live on.

Great work! You now know how to:

• produce a separation to show a space is disconnected;
• use a clopen subset as the same evidence in a different form;
• cut the rationals at an irrational point;
• see why the Sierpiński space is connected.