T.5 · Open sets, compactness and connectedness

In a metric space, an open set contains a small ball around each of its points. The interior consists of such points; the closure includes all limit points; the boundary is closure minus interior. Open and closed are not mutually exclusive labels; the empty set and the whole space are both. In R, the set [0,1) has interior (0,1), closure [0,1] and boundary {0,1}.

In a subspace X, an open set has the form X intersected with an ambient open set. Thus [0,1) is open relative to [0,2], using intersection with (-1,1), although it is not open in R. A set may also be relatively closed without being closed in the ambient space. Always state which space defines neighbourhoods and which metric is used.

Compactness means every open cover has a finite subcover. In Euclidean R^n, Heine–Borel makes this equivalent to closed and bounded. In a metric space, compactness is equivalent to sequential compactness; completeness and boundedness alone do not suffice in arbitrary metric spaces. A continuous image of a compact set is compact, giving attained maxima and minima for real continuous functions on a nonempty compact domain.

Connected sets cannot be separated into two disjoint nonempty relatively open parts; connected subsets of R are precisely intervals. A continuous image of a connected set is connected, which yields the intermediate value theorem. Path connectedness implies connectedness, but not conversely in every space. A compact set need not be connected, and a connected set need not be compact; a finite two-point set and an open interval supply the contrasting cases.

Let X=[0,1] with its usual relative topology. The set U=[0,0.5) equals X intersected with (-1,0.5), so U is open in X. Its closure in X is [0,0.5], and its boundary in X is {0.5}; zero is an interior point relative to X. For a continuous f on X with f(0)<0<f(1), connectedness ensures a zero, while compactness separately ensures attained extrema.

Heine–Borel's closed-and-bounded test is a Euclidean-space theorem. Connectedness and compactness answer different questions.

In $X=[0,2]$ with its relative topology, find the interior, closure and boundary of $A=[0,1)$. Is A open or closed in X?

Worked solution

$A=X\cap(-1,1)$ is open in X, so its relative interior is A. Its closure is $[0,1]$ and its boundary is $\{1\}$. Zero is interior relative to X because a sufficiently small ball in X has no negative part. A is not closed in X: the sequence $1-1/n$ for $n\ge2$ tends to the missing point 1. In the ambient real line, interior and boundary would instead be $(0,1)$ and $\{0,1\}$.

Give a compact disconnected real set and a connected noncompact real set. Must a continuous real function on the latter attain an extreme value? Explain which property gives the intermediate value theorem.

Check after attempting

The finite set $\{0,2\}$ is compact but disconnected. The interval $(0,1)$ is connected but not compact. The continuous identity function on $(0,1)$ attains neither a maximum nor a minimum. Connectedness makes a continuous real image an interval and therefore supplies intermediate values. Compactness instead makes a nonempty continuous real image attain its extreme values. Neither property substitutes for the other.