A useful embedding for Cp(X)

Let $X$ be a Tychonoff space (also called completely regular space). By $C_p(X)$ we mean the space of all continuous real-valued functions defined on $X$ endowed with the pointwise convergence topology. In this post we discuss a scenario in which a function space can be embedded into another function space. We prove the following theorem. An example follows the proof.

Theorem 1
Suppose that the space $Y$ is a continuous image of the space $X$. Then $C_p(Y)$ can be embedded into $C_p(X)$.

Proof of Theorem 1
Let $t:X \rightarrow Y$ be a continuous surjection, i.e., $t$ is a continuous function from $X$ onto $Y$. Define the map $\psi: C_p(Y) \rightarrow C_p(X)$ by $\psi(f)=f \circ t$ for all $f \in C_p(Y)$. We show that $\psi$ is a homeomorphism from $C_p(Y)$ into $C_p(X)$.

First we show $\psi$ is a one-to-one map. Let $f,g \in C_p(Y)$ with $f \ne g$. There exists some $y \in Y$ such that $f(y) \ne g(y)$. Choose some $x \in X$ such that $t(x)=y$. Then $f \circ t \ne g \circ t$ since $(f \circ t)(x)=f(t(x))=f(y)$ and $(g \circ t)(x)=g(t(x))=g(y)$.

Next we show that $\psi$ is continuous. Let $f \in C_p(Y)$. Let $U$ be open in $C_p(X)$ with $\psi(f) \in U$ such that $U=\left\{q \in C_p(X): \forall \ i=1,\cdots,n, \ q(x_i) \in U_i \right\}$

where $x_1,\cdots,x_n$ are arbitrary points of $X$ and each $U_i$ is an open interval of the real line $\mathbb{R}$. Note that for each $i$, $f(t(x_i)) \in U_i$. Now consider the open set $V$ defined by: $V=\left\{r \in C_p(Y): \forall \ i=1,\cdots,n, \ r(t(x_i)) \in U_i \right\}$

Clearly $f \in V$. It follows that $\psi(V) \subset U$ since for each $r \in V$, it is clear that $\psi(r)=r \circ t \in U$.

Now we show that $\psi^{-1}: \psi(C_p(Y)) \rightarrow C_p(Y)$ is continuous. Let $\psi(f)=f \circ t \in \psi(C_p(Y))$ where $f \in C_p(Y)$. Let $G$ be open with $\psi^{-1}(f \circ t)=f \in G$ such that $G=\left\{r \in C_p(Y): \forall \ i=1,\cdots,m, \ r(y_i) \in G_i \right\}$

where $y_1,\cdots,y_m$ are arbitrary points of $Y$ and each $G_i$ is an open interval of $\mathbb{R}$. Choose $x_1,\cdots,x_m \in X$ such that $t(x_i)=y_i$ for each $i$. We have $f(t(x_i)) \in G_i$ for each $i$. Define the open set $H$ by: $H=\left\{q \in \psi(C_p(Y)) \subset C_p(X): \forall \ i=1,\cdots,m, \ q(x_i) \in G_i \right\}$

Clearly $f \circ t \in H$. Note that $\psi^{-1}(H) \subset G$. To see this, let $r \circ t \in H$ where $r \in C_p(Y)$. Now $r(t(x_i))=r(y_i) \in G_i$ for each $i$. Thus $\psi^{-1}(r \circ t)=r \in G$. It follows that $\psi^{-1}$ is continuous. The proof of the theorem is now complete. $\blacksquare$

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Example

The proof of Theorem 1 is not difficult. It is a matter of notating carefully the open sets in both function spaces. However, the embedding makes it easy in some cases to understand certain function spaces and in some cases to relate certain function spaces.

Let $\omega_1$ be the first uncountable ordinal, and let $\omega_1+1$ be the successor ordinal to $\omega_1$. Furthermore consider these ordinals as topological spaces endowed with the order topology. As an application of Theorem 1, we show that $C_p(\omega_1+1)$ can be embedded as a subspace of $C_p(\omega_1)$. Define a continuous surjection $g:\omega_1 \rightarrow \omega_1+1$ as follows: $g(\gamma) = \begin{cases} \omega_1 & \mbox{if } \ \gamma =0 \\ \gamma-1 & \mbox{if } \ 1 \le \gamma < \omega \\ \gamma & \mbox{if } \ \omega \le \gamma < \omega_1 \end{cases}$

The map $g$ is continuous from $\omega_1$ onto $\omega_1+1$. By Theorem 1, $C_p(\omega_1+1)$ can be embedded as a subspace of $C_p(\omega_1)$. On the other hand, $C_p(\omega_1)$ cannot be embedded in $C_p(\omega_1+1)$. The function space $C_p(\omega_1+1)$ is a Frechet-Urysohn space, which is a property that is carried over to any subspace. The function $C_p(\omega_1)$ is not Frechet-Urysohn. Thus $C_p(\omega_1)$ cannot be embedded in $C_p(\omega_1+1)$. A further comparison of these two function spaces is found in this subsequent post.

____________________________________________________________________ $\copyright \ 2014 \text{ by Dan Ma}$