A subset is connected when it is connected as a subspace, and as with compactness the test can be written with open sets of the surrounding space instead of traces. Two facts then do most of the work in this chapter: connected pieces sharing a point unite into a connected whole, and the connected subsets of the line are exactly the intervals.

Two fields joined by a shared gate make one field. Two fields with a road between them do not, however close they lie.

The subset $A=[-3,-1]\cup[1,3]$ of the line, with the point $0$ missing from it. The rays $(-\infty,0)$ and $(0,\infty)$ are open in $\mathbb{R}$, each meets $A$, together they cover $A$, and they share no point — a separation of $A$. Any subset of the line that misses a point lying between two of its own comes apart the same way, which is why the connected subsets of the line are the intervals.

A subset $A\subseteq X$ is connected when the subspace $A$ has no separation. Written with open sets of $X$: there are no open $U,V$ of $X$ with $A\subseteq U\cup V$, both $A\cap U$ and $A\cap V$ non-empty, and $A\cap U\cap V=\varnothing$. Note that $U$ and $V$ may meet each other outside $A$; it is only inside $A$ that they must not.

• Connected subsets sharing at least one point unite into a connected subset, however many of them there are. This is the theorem below, and it is the tool the rest of the chapter uses.
• The connected subsets of $\mathbb{R}$ are exactly the intervals — including single points, rays and the whole line. One direction is proved below; that an interval is connected rests on the least upper bound property and is taken on trust, as it has been since this chapter opened.
• If $A$ is connected and $A\subseteq B\subseteq\overline{A}$, then $B$ is connected: adding accumulation points of a connected set cannot break it. Stated here and used for components; the proof is a short argument about traces and is left to a later course.
• A subset can be connected in itself while the space around it is not, and disconnected while the space is connected: $[0,1]$ is connected inside the disconnected $[0,1]\cup[2,3]$, and $\{0\}\cup\{1\}$ is disconnected inside the connected .
• $U$ and $V$ are allowed to meet away from $A$. Demanding disjoint open sets of the whole space is a stronger condition, and in a general space the two demands differ.

Suppose the union were separated by two open sets. The shared point lies in one of them, say the first. Take any one of the pieces: it lies inside the union, so the same two open sets cover it, and they share no point of it. That makes them a separation of the piece — unless one of the two meets it in nothing. The piece is connected, so one of them must miss it entirely, and since the piece holds the shared point, which is in the first, the one that misses it is the second. This is true of every piece, so the second open set meets no piece at all, and the pieces are everything. So it is empty, and a separation needs both halves non-empty.

Suppose the union came apart, with the shared point in the first open set.

The same two open sets cover each piece and share no point of it.

Each piece is connected, so it cannot be met by both: otherwise the pair would separate it.

Every piece holds the shared point, which lies in the first set, so it is the second that misses each piece.

The second set therefore meets nothing in the union, so it was not a separation after all.

The two open sets come from the whole space; they must cover $A$, meet it both, and share no point of it.

• A subset is tested with open sets of the space around it, disjoint only inside the subset.
• Connected pieces sharing a point unite into a connected set.
• The connected subsets of the line are exactly the intervals.
• Closures of connected sets are connected.

Great work! You now know how to:

• test a subset for connectedness with open sets of the space;
• unite connected sets that share a point;
• recognise the connected subsets of the line as the intervals;
• use the closure of a connected set as connected.