The Suslinian number and other cardinal invariants of continua

The simplest examples of non-metrizable Suslinian continua are Suslin lines. However this class of examples has a consistency flavour since no Suslin line exists in some models of ZFC (for example, in models satisfying (MA+¬CH) ). It turns out that any example of a non-metrizable locally connected Suslinian continuum necessarily has consistency nature: the existence of such a continuum is equivalent to the existence of a Suslin line, see [1]. This implies that under the Suslin Hypothesis (asserting that no Suslin line exists) each locally connected Suslinian continuum is metrizable.

It is clear that each Suslinian continuum X has countable Suslin number c(X). At this point we recall the definition of some known topological cardinal invariants. Given a topological space X let

• c(X) = sup{|U| : U is a disjoint family of non-empty open subsets of X} be the Suslin number of X; • l(X) = min{κ : each open cover of X contains a subcover of size ≤ κ} be the Lindelöf number of X; • d(X) = min{|D| : D is a dense set in X} be the density of X;

• hl(X) = sup{l(Y ) : Y ⊂ X} be the hereditary Lindelöf number of X;

• hd(X) = sup{d(Y ) : Y ⊂ X} be the hereditary density of X;

• w(X) = min{|B| : B is a base of the topology of X} be the weight of X;

• rim-w(X) = min{sup U∈B w(∂U ) : B is a base of the topology of X} be the rim-weight of X. In the context of Suslinian continua, by analogy with the Suslin number c(X) it is natural to introduce a new cardinal invariant

• Sln(X) = sup{|C| : C is a disjoint family of non-degenerate subcontinua of X} defined for any continuum X and called the Suslinian number of X. Thus a continuum X is Suslinian if and only Sln(X) ≤ ℵ 0 .

It is clear that Sln(X) ⊂ Sln(Y ) for any pair X ⊂ Y of continua. It will be convenient to extend the definition of Sln(X) to all Tychonov spaces letting Sln(X) = min{Sln(Y ) : Y is a continuum containing X} for a Tychonov space X.

Like many other cardinal invariants the Suslinian number is monotone.

Proposition 1. If X is a Tychonov space and Y is a subspace of X, then Sln(Y ) ≤ Sln(X).

The cardinal invariant Sln(X) is not trivial since it can attain any infinite value.

Note that the each hedgehog is rim-finite in the sense that it has a base of the topology consisting of sets with finite boundaries. Let us remark that a rim-finite continuum X with uncountable Suslinian number must be non-metrizable (because rim-countable metrizable continua are Suslinian, see [6]).

The Suslinian number can not increase under monotone maps. We recall that a map f :

Proof. Embed X in a continuum Z with Sln(Z) = Sln(X). Consider the following equivalence relation on Z: x ∼ y if either x = y or x, y ∈ X and f (x) = f (y). Let T = Z/ ∼ be the quotient space and q : Z → T be the quotient map. Since all the equivalence classes are connected, the quotient map q is monotone. Since the preimage of a connected set under a monotone map is connected, Sln(T ) ≤ Sln(Z). It remains to observe that Y can be identified with a subspace of T , wich yields Sln(Y ) ≤ Sln(T ) ≤ Sln(Z) = Sln(X).

In the subsequent proof we shall refer to properties of the hyperspace exp(X) of a given compact Hausdorff space X. The hyperspace exp(X) of X is the space of all non-empty closed subsets of X, endowed with the Vietoris topology. It is well known that exp(X) is a compact Hausdorff space with w(exp(X)) = w(X). By exp c (X) we denote the subspace of exp(X) consisting of subcontinua of X. It is easy to see that exp c (X) is a closed subspace in exp(X).

Recall that a map f :

Proof. Let Z be a continuum such that Z ⊃ X and Sln(X) = Sln(Z). Embed Y into the Tychonov cube [0, 1] κ where κ = w(Y ). It follows from the Tietze-Urysohn Theorem that the map f can be extended to a map f : Z → [0, 1] κ . Observe that each non-empty open set U ⊂ Z has no one-point component (otherwise this one-point component would be a quasi-component and consequently Z would contain a non-trivial clopen subset which contradicts the connectedness of Z). Each component of U contains a non-trivial subcontinuum and consequently, U has at most Sln(X) components. Denote by C U the family of closures of components of U .

Let B be a base for the topology of f (Z) with |B| ≤ κ. Finall

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