# A useful representation of Cp(X)

Let $X$ be a completely regular space. The space $C_p(X)$ is the space of all real-valued continuous functions defined on $X$ endowed with the pointwise convergence topology. In this post, we show that $C_p(X)$ can be represented as the product of a subspace of $C_p(X)$ with the real line $\mathbb{R}$. We prove the following theorem. See here for an application of this theorem.

Theorem 1
Let $X$ be a completely regular space. Let $x \in X$. Let $Y$ be defined by:

$Y=\left\{f \in C_p(X): f(x)=0 \right\}$

Then $C_p(X)$ is homeomorphic to $Y \times \mathbb{R}$.

The above theorem can be found in [1] (see Theorem I.5.4 on p. 37). In [1], the homeomorphism is stated without proof. For the sake of completeness, we provide a detailed proof of Theorem 1.

Proof of Theorem 1
Define $h: C_p(X) \rightarrow Y \times \mathbb{R}$ by $h(f)=(f-f(x),f(x))$ for any $f \in C_p(X)$. The map $h$ is a homeomorphism.

The map is one-to-one

First, we show that it is a one-to-one map. Let $f,g \in C_p(X)$ where $f \ne g$. Assume that $f(x) \ne g(x)$. Then $h(f) \ne h(g)$. So assume that $f(x)=g(x)$. Then the functions $f-f(x)$ and $g-g(x)$ are different, which means $h(f) \ne h(g)$.

The map is onto

Now we show $h$ maps $C_p(X)$ onto $Y \times \mathbb{R}$. Let $(g,t) \in Y \times \mathbb{R}$. Let $f=g+t$. Note that $f(x)=g(x)+t=t$. Then $f-f(x)=g$. We have $h(f)=(g,t)$.

Note. Showing the continuity of $h$ and $h^{-1}$ is a matter of working with the basic open sets in the function space carefully (e.g. making the necessary shifting). Some authors just skip the details and declare them continuous, e.g. [1]. Readers are welcome to work out enough of the details to see the key idea.

The map is continuous

Show that $h$ is continuous. Let $f \in C_p(X)$. Let $U \times V$ be an open set in $Y \times \mathbb{R}$ such that $h(f) \in U \times V$ and,

$U=\left\{g \in Y: \forall \ i=1,\cdots,n, g(x_i) \in U_i \right\}$

$\forall \ i=1,\cdots,n, \ U_i=(f(x_i)-f(x)-\frac{1}{k},f(x_i)-f(x)+\frac{1}{k})$

$V=(f(x)-\frac{1}{k},f(x)+\frac{1}{k})$

where $x_1,\cdots,x_n$ are arbitrary points in $X$ and $k$ is some large positive integer. Define the following:

$\forall \ i=1,\cdots,n, \ W_i=(f(x_i)-\frac{1}{2k},f(x_i)+\frac{1}{2k})$

$W_{n+1}=(f(x)-\frac{1}{2k},f(x)+\frac{1}{2k})$

$x_{n+1}=x$

Then define the open set $W$ as follows:

$W=\left\{q \in C_p(X): \forall \ i=1,\cdots,n,n+1, q(x_i) \in W_i \right\}$

Clearly $f \in W$. We need to show $h(W) \subset U \times V$. Let $q \in W$. Then $h(q)=(q-q(x),q(x))$. We need to show that $q-q(x) \in U$ and $q(x) \in V$. Note that $q(x_{n+1})=q(x) \in W_{n+1}$. For each $i=1,\cdots,n$, $q(x_i) \in W_i$. So we have the following:

$f(x_i)-\frac{1}{2k}

$f(x)-\frac{1}{2k}

Subtracting the above two inequalities, we have the following:

$f(x_i)-f(x)-\frac{1}{k}

The above inequality shows that for each $i=1,\cdots,n$, $q(x_i) -q(x) \in U_i$. Hence $q-q(x) \in U$. It is clear that $q(x) \in V$. This completes the proof that the map $h$ is continuous.

The inverse is continuous

We now show that $h^{-1}$ is continuous. Let $(g,t) \in Y \times \mathbb{R}$. Note that $h^{-1}(g,t)=g+t$. Let $M$ be an open set in $C_p(X)$ such that $g+t \in M$ and

$M=\left\{f \in C_p(X): \forall \ i=1,\cdots,n+1, f(x_i) \in M_i \right\}$

$\forall \ i=1,\cdots,n, \ M_i=(g(x_i)+t-\frac{1}{m},g(x_i)+t+\frac{1}{m})$

$x_{n+1}=x$

$M_{n+1}=(t-\frac{1}{m},t+\frac{1}{m})$

where $x_1,\cdots,x_n$ are arbitrary points of $X$ and $m$ is some large positive integer. Now define an open subset $G \times T$ of $Y \times \mathbb{R}$ such that $(g,t) \in G \times T$ and

$G=\left\{q \in Y: \forall \ i=1,\cdots,n+1, q(x_i) \in G_i \right\}$

$\forall \ i=1,\cdots,n, \ G_i=(g(x_i)-\frac{1}{2m},g(x_i)+\frac{1}{2m})$

$T=(t-\frac{1}{2m},t+\frac{1}{2m})$

We need to show that $h^{-1}(G \times T) \subset M$. Let $(q,a) \in G \times T$. We then have the following inequalities.

$\forall \ i=1,\cdots,n, \ g(x_i)-\frac{1}{2m}

$t-\frac{1}{2m}

Adding the above two inequalities, we obtain:

$\forall \ i=1,\cdots,n, \ g(x_i)+t-\frac{1}{m}

The above implies that $\forall \ i=1,\cdots,n$, $q(x_i)+a \in M_i$. It is clear that $q(x_{n+1})+a=q(x)+a=a \in M_{n+1}$. Thus $q+a \in M$. This completes the proof that $h^{-1}$ is continuous.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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