One point compactification

Theorem (One point compactification) Any locally compact space (X) can be embedded in another compact space (Y), which just has one more point than (X), such that the relative topology of (X) with respect to (Y) is the same as the original topology of (X). The space (Y) thus constructed is called one point compactification of (X).

Proof The key step of the proof is to construct the topology of (Y) by adopting the open sets from (X) whose complements are compact in (X).

Construction of the topology (mathcal{T}_Y) for (Y)

Let ({U_i}_{i in I}) be a collection of open sets in (X) such that (X - U_i) is compact in (X) for all (i in I). Figuratively speaking, these open sets are the spaces (X) punctured with holes having bounded dimension. We also note that because the space (X) is locally compact, hence for all (x in X), there is a compact neighborhood containing (x). This ensures the above selected collection of open sets is not empty.

Let (Y) be the space by appending one point (y_0) to (X). Hence, (Y = X cup {y_0}). Let (mathcal{T}_Y) be the topology of (Y) which is defined via the following two rules.

For all (V in mathcal{T}_Y),

1. if (V) does not contain (y_0), (V) is an open set of (X) in (X)'s original topology (mathcal{T}_X).
2. if (V) contains (y_0), then (V cap X in {U_i}_{i in I}), where (X - U_i) is compact in (X).

Prove (mathcal{T}_Y) is a topology of (Y)

We'll then prove that such constructed (mathcal{T}_Y) really defines a topology for (Y).

• The empty set (Phi) does not contain (y_0) and it belongs to (mathcal{T}_X), so (Phi in mathcal{T}_Y) according to rule 1.
• The whole space (Y) contains the point (y_0) and (Y cap X = X). Because (X^{c} = Phi), which is compact in (X) as a trivial case, (Y) belongs to (mathcal{T}_Y).

• We should check that: a) the union of any number of sets in (mathcal{T}_Y) belongs to (mathcal{T}_Y); b) the intersection of any finite number of sets in (mathcal{T}_Y) belongs to (mathcal{T}_Y).

  • If the selected sets from (mathcal{T}_Y) all satisfy rule 1, i.e., they are selected from (X)'s original topology (mathcal{T}_X), it is obvious that the union and finite intersection of these sets are still open in (X). Because they do not contain (y_0), they satisfy rule 1.

  • If the selected sets from (mathcal{T}_Y) all satisfy rule 2, they all contain point (y_0). Let such selected collection of sets be ({V_j}_{j in J}). Let ({U_j}_{j in J}) be the associated open sets such that (U_j = V_j cap X) for all (j in J). Then we have

    [
    left( left( igcup_{j in J} V_j ight) cap X ight) ^c = left( igcup_{i in J} U_j ight)^c = igcap_{j in J} U_j^c.
    ]

    Because all (U_j^c) are compact and hence closed in (X), their intersection is also closed. Because a closed subset of a compact set is still compact, the intersection of all (U_j^c) is compact. Thus, rule 2 is satisfied.

    Similarly, the intersection of a finite number of sets selected from (mathcal{T}_Y) satisfying rule 2 can be proved to belong to (mathcal{T}_Y).

  • If there are two subgroups in the selected sets from (mathcal{T}_Y), such that the sets in one subgroup satisfy rule 1 and those in the other subgroup satisfy rule 2, the union or intersection of all the selected sets can also be divided into two groups corresponding to the two rules. Then we only need to verify:

    (forall V_1, V_2 in mathcal{T}_Y), where (V_1) satisfies rule 1 and (V_2) satisfies rule 2, then both (V_1 cup V_2) and (V_1 cap V_2) are in (mathcal{T}_Y).

    For the union of (V_1) and (V_2), it contains (y_0). Then we check if the complement of ((V_1 cup V_2) cap X) is compact in (X).

    [
    left( V_1 cup V_2 ight) cap X = (V_1 cap X) cup (V_2 cap X) = V_1 cup U_2.
    ]

    Then

    [
    (V_1 cup U_2)^c = V_1^c cap U_2^c,
    ]

    which is a closed subset of the compact set (U_2^c). Therefore, the complement of ((V_1 cup V_2) cap X) is compact.

    For the intersection of (V_1) and (V_2), it does not contain (y_0). We have

    [
    V_1 cap V_2 = V_1 cap U_2.
    ]

    Because both (V_1) and (U_2) are open sets in (X), their intersection is an open set in X, so is (V_1 cap V_2).

Summarizing the above, we've proved (mathcal{T}_Y) is really a topology for (Y). It is also obvious to see from the above proof that the relative topology of (X) with respect to (Y) is the same as its original topology.

Prove (Y) is compact

Let ({V_i}_{i in I}) be an open covering of (Y). Then ({ V_i cap X }_{i in I}) is an open covering of (X). Meanwhile, there exists an index (i_0 in I) such that (y_0 in V_{i_0}). Let (U_{i_0} = V_{i_0} cap X), so (U_{i_0}^c) is compact in (X). Because ({ V_i cap X }_{i in I}) covers (U_{i_0}^c), there exists a finite subcovering

[
U_{i_0}^c subset igcup_{k=1}^n { V_{i_k} cap X }.
]

Because (V_{i_0}) contains both (y_0) and (U_{i_0}), the collection ({ V_{i_k}}_{k=1}^n) appended with (V_{i_0}) forms an finite subcovering of (Y). Hence (Y) is compact.

Examples of one point compactification

1. The real line (mathbb{R}) is not compact. By adding an infinity point (infty), the real line can be transformed to a circle with (infty) as the paste point.
2. The plane (mathbb{R}^2) is not compact. By adding an infinity point (infty), the plane can be transformed to a sphere with (infty) as the paste point.