Question

Problem:

Part A: Let X be a topological space and ∞ / ∈ X. Define b X = X ∪{∞} and consider on b X the family: T = {A ⊂ X| A is open in X}[nb X \K| K is closed and compact in Xo. 1. Show that (b X,T) is a compact topological space. It is henceforth called a one-point compactification of X. 2. Show that b X is Hausdorff if and only if X is Hausdorff and every point in X has a compact neighborhood. 3. Show that a compact Hausdorff space Y coincides with a one-point compactification of Y \{y} ∀y ∈ Y . 4. Show that if f : X → Y is a continuous map such that the pre-image f−1(K) of any compact K ⊂ Y is a compact in X, f extends naturally to a continuous mapb f : b X →b Y .

Answer 1

$(X,\tau)\text{ --- topological space}.\\ \text{One-point compactification.}\\ \infty\notin X\\ X^*:=X\cup\{\infty\}\\ \text{Topology }\tau^* \text{ on }X^*:\\ \tau^*=\tau\cup\{V\sube X^*|\infty\in V\text{ and }X\diagdown V\text{ --- closed and compact in }X\}.\\ \text{We denote }(X^*,\tau^*) \text{ by } \sigma(X).\\ 1.\text{ Show that }\sigma(x)\text{ --- compactification of X}.\\ (X,\tau)\text{ is not compact}.\\ \text{First we will show that }\sigma(X)\text{ is compact. Let } U\text{ be an open cover of }\sigma(X).\\ \text{We take some }V\text{ that contains }\infty.\text{ By definition of }\tau^*,\ X^*\diagdown V \text{ is a compact}\\ \text{subset of }X\text{ which is covered by }U \diagdown\{V\}. \text{ So we have some finite subcover of}\\ X\diagdown V\text{ which with the addition of } V\text{ is a finite subcover of }\sigma(X).\\ \overline{X}=X^* \text{ since the the only way }\{\infty\}\text{ can be open in }\sigma(X)\text{ is if }X\text{ is compact}.\\ i: X\rightarrow X^* \text{ given by }\\ i(x)=x\\ \text{is an embedding (since the open subsets of } X\text{ in } \sigma(X)\text{ are exactly}\\ \text{the elements of }\tau).\\ 2.\ \Rightarrow\\ \text{We take }x,\ y\in X^*,\ x\neq y.\\ \text{If }x,\ y \in X,\text{ then we can separate them with open sets since }X \text{ is Hausdorff}.\\ \text{We assume }y=\infty\text{ without loss of generality}.\\ X\text{ is locally compact. So we can find a compact set }K\sube X\text{ and an open set }\\ U\sube X\text{ such that } x\in U\sube K.\ U \text{ and } X^*\diagdown K \text{ are disjoint}\\ \text{open subsets of }X^* \text{ separating } x \text{ and } y, \text{respectively.}$

$\Leftarrow$

Suppose $\sigma(X)$ is Hausdorff. Then $X$ is Hausdorff since it is homeomorphic to a subspace of a Hausdorff space. $X$ is locally compact since it is homeomorphic to an open subset of the locally compact space $\sigma(X)$.

$3.$

A compactification of a topological space $X$ is a pair $(Y;f)$ consisting of a topological space $Y$ and a continuous map $f: X\longrightarrow Y$ such that: $Y$ is compact, $f$ is a homeomorphism onto $f(X)$, and $f(X)$ is dense in $Y$, i. e. $\overline{f(X)}=Y$.

In our case $f: Y\diagdown \{y\}\longrightarrow Y$.

$Y$ is compact.

$f$ is a homeomorphism onto $f(Y\diagdown\{y\}).$

$\overline{f(Y\diagdown\{y\})}=Y.$

$4.$

Show that if $f: X\longrightarrow Y$ is a continuous map such that the pre-image $f^{-1}(K)$ of any compact $K\subset Y$ is a compact in $X$, $f$ extends naturally to a continuous map$\sigma(f):\sigma(X)\longrightarrow \sigma(Y).$

We should define $\sigma(f)(x)=f(x)$ for $x\neq\infty$ and $\sigma(f)(\infty)=\infty$.

Let $U\subset Y$ be an open subset.

If $\infty\notin U$ then $(\sigma(f))^{-1}(U)=f^{-1}(U)$ is open since $f$ is continuous.

If $\infty\in U$ then $U=(Y\diagdown K)\cup\{\infty\}$ and $(\sigma(f))^{-1}((Y\diagdown K)\cup\{\infty\})=(\sigma(f))^{-1}(Y\diagdown K)\cup(\sigma(f))^{-1}(\{\infty\})=(\sigma(f))^{-1}(Y\diagdown K)\cup\{\infty\}.\\ (\sigma(f))^{-1}(Y\diagdown K)=X\diagdown (\sigma(f))^{-1}(K).$

$K$ is closed and compact.