One theorem about normality of Cp(X)

Assuming that the function space C_p(X) is normal, what can be said about the domain space X? In this post, we prove a theorem that yields a corollary that for any normal space X, if C_p(X) is normal, then X has countable extent (i.e. every closed and discrete subset of X is countable). Thus the normality of the function space limits the size of a closed and discrete subset of the domain space. It then follows that for any metric space X, if C_p(X) is normal, X has is second countable (i.e. having a countable base). Another immediate, but slightly less obvious, corollary is that for any X that is a normal Moore space, if C_p(X) is normal, then X is metrizable.

For definitions of basic open sets and other background information on the function space C_p(X), see this previous post.

Let X be a space. Let Y \subset X. Let \pi_Y be the natural projection from the product space \mathbb{R}^X into the product space \mathbb{R}^Y. Specifically, if f \in \mathbb{R}^X, then \pi_Y(f)=f \upharpoonright Y, i.e., the function f restricted to Y. In the discussion below, \pi_Y is defined just on C_p(X), i.e., \pi_Y is the natural projection from C_p(X) into C_p(Y). It is always the case that \pi_Y(C_p(X)) \subset C_p(Y). It is not necessarily the case that \pi_Y(C_p(X))=C_p(Y). However, if X is a normal space and Y is closed in X, then \pi_Y(C_p(X))=C_p(Y) and \pi_Y is the natural projection from C_p(X) onto C_p(Y). We prove the following theorem.

Theorem 1

    Suppose that C_p(X) is a normal space. Let Y be a closed subspace of X. Then \pi_Y(C_p(X)) is a normal space.

Theorem 1 is found in [1] (see Theorem I.6.2). In proving Theorem 1, we need the following lemma.

Lemma 2

    Let T=\prod_{\alpha \in A} T_\alpha be a product of separable metrizable spaces. Let S be a dense subspace of T. Then the following conditions are equivalent.

    1. S is normal.
    2. For any pair of disjoint closed subsets H and K of S, there exists a countable B \subset A such that \overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing.
    3. For any pair of disjoint closed subsets H and K of S, there exists a countable B \subset A such that \pi_B(H) and \pi_B(K) are separated in \pi_B(S), meaning that \overline{\pi_B(H)} \cap \pi_B(K)=\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing.

For a proof of Lemma 2, see Lemma 1 in this previous post.

Proof of Theorem 1
Note that \pi_Y(C_p(X)) is a dense subspace of \mathbb{R}^Y. Let H and K be disjoint closed subsets of \pi_Y(C_p(X)). To show \pi_Y(C_p(X)) is normal, by Lemma 2, we only need to produce a countable B \subset Y such that \overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing. The closure here is taken in \pi_B(\pi_Y(C_p(X))).

Let H_1=\pi_Y^{-1}(H) and K_1=\pi_Y^{-1}(K). Both H_1 and K_1 are closed subsets of C_p(X). By Lemma 2, there exists some countable C \subset X such that \overline{\pi_C(H_1)} \cap \overline{\pi_C(K_1)}=\varnothing. The closure here is taken in \pi_C(C_p(X)). According to the remark at the end of this previous post, for any countable D \subset X such that C \subset D, \overline{\pi_D(H_1)} \cap \overline{\pi_D(K_1)}=\varnothing. In other words, the countable set C can be enlarged and the conclusion of the lemma still holds. With this observation in mind, we can assume that C \cap Y \ne \varnothing. If not, we can always throw countably many points of Y into C and still have \overline{\pi_C(H_1)} \cap \overline{\pi_C(K_1)}=\varnothing.

Let B=C \cap Y. We claim that \overline{\pi_B(H_1)} \cap \overline{\pi_B(K_1)}=\varnothing. The closure here is taken \pi_B(C_p(X)). Suppose that \overline{\pi_B(H_1)} \cap \overline{\pi_B(K_1)} \ne \varnothing. Choose f \in C_p(X) such that f \upharpoonright B \in \overline{\pi_B(H_1)} \cap \overline{\pi_B(K_1)}. It follows that f \upharpoonright C \in \overline{\pi_C(H_1)}. To see this, let f \upharpoonright C \in U=\prod_{x \in C} U_x where U is a standard basic open set. Let F be the support of U, i.e., the finite set of x \in C such that U_x \ne \mathbb{R}. Let F_1=F \cap Y and F_2=F \cap (X-Y). Let U^*=\prod_{x \in B} U_x. Note that f \upharpoonright B \in U^*. Since f \upharpoonright B \in \overline{\pi_B(H_1)}, there is some g \in H_1 such that \pi_B(g) \in U^*. Note that F_1 is the support of U^*.

Because the space X is completely regular, there is a h \in C_p(X) such that h(x)=0 for all x \in Y and h(x)=f(x)-g(x) for all x \in F_2. Let w=h+g. Since w \upharpoonright Y=g \upharpoonright Y, w \in H_1. Note that w=g on Y, hence on F_1 and that w=f on F_2. Thus w \upharpoonright C \in U. Since U is an arbitrary open set containing f \upharpoonright C, it follows that f \upharpoonright C \in \overline{\pi_C(H_1)}. By a similar argument, it can be shown that f \upharpoonright C \in \overline{\pi_C(K_1)}. This is a contradiction since \overline{\pi_C(H_1)} \cap \overline{\pi_C(K_1)}=\varnothing. Therefore the claim that \overline{\pi_B(H_1)} \cap \overline{\pi_B(K_1)}=\varnothing is true, with the closure being taken in \pi_B(C_p(X)).

Because B \subset Y, observe that \pi_B(H_1)=\pi_B(H) and \pi_B(K_1)=\pi_B(K). Furthermore, \pi_B(\pi_Y(C_p(X)))=\pi_B(C_p(X)). Thus we can claim that \overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing, with the closure being taken in \pi_B(\pi_Y(C_p(X))). By Lemma 2, \pi_Y(C_p(X)) is normal. \blacksquare

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Some Corollaries

Corollary 3

    Let X be a normal space. If C_p(X) is normal, then X has countable extent, i.e., every closed and discrete subset of X is countable.

Proof of Corollary 3
Let Y be a closed and discrete subset of X. We show that Y must be countable. Since Y is closed and X is normal, \pi_Y(C_p(X))=C_p(Y). By Theorem 1, C_p(Y) is normal. Since Y is discrete, C_p(Y)=\mathbb{R}^Y. If Y is uncountable, \mathbb{R}^Y is not normal. Thus Y must be countable. \blacksquare

Corollary 4

    Let X be a metrizable space. If C_p(X) is normal, then X has a countable base.

Proof of Corollary 4
Note that in any metrizable space, the weight equals the extent. By Corollary 3, X has countable extent and thus has countable base. \blacksquare

Corollary 5

    Let X be a normal space. If C_p(X) is normal, then X is collectionwise normal.

Proof of Corollary 5
Any normal space with countable extent is collectionwise normal. See Theorem 2 in this previous post. \blacksquare

Corollary 6

    Let X be a normal Moore space. If C_p(X) is normal, then X is metrizable.

Proof of Corollary 6
Suppose C_p(X) is normal. By Theorem 1, X has countable extent. By Corollary 5, X is collectionwise normal. According to Bing’s metrization theorem, any collectionwise normal Moore space is metrizable (see [2] Theorem 5.4.1 in page 329). \blacksquare

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Reference

  1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
  2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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\copyright \ 2014 \text{ by Dan Ma}

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