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}<q(x_i)<f(x_i)+\frac{1}{2k}

    f(x)-\frac{1}{2k}<q(x)<f(x)+\frac{1}{2k}

Subtracting the above two inequalities, we have the following:

    f(x_i)-f(x)-\frac{1}{k}<q(x_i)-q(x)<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}<q(x_i)<g(x_i)+\frac{1}{2m}

    t-\frac{1}{2m}<a<t+\frac{1}{2m}

Adding the above two inequalities, we obtain:

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

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One thought on “A useful representation of Cp(X)

  1. Pingback: Normality in Cp(X) | Dan Ma's Topology Blog

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