Let M be a metric space. Let T(M) be the topology of M (i.e. the collection of all open subsets of M) and let C(M) be the collection of all connected subsets of M. In my opinion one often has a much clearer intuitive picture of the topology of a metric space M if one knows C(M), than if one knows T(M). I would be interested in characterizing the largest collection S of metric spaces M for which T(M) is uniquely determined when C(M) is specified. To show that all Banach spaces belong to S, let the term "*continuum" denote any non-empty closed and connected (but not necessarily compact) subset of a metric space. We can define *continua in terms of connected sets by saying that a *continuum is either identical with the metric space M (if M is connected) or else it is a non-empty proper subset Q of M such that the union of Q and the singleton of p is not connected, whenever p is a point of M that does not belong to Q. Now if B is a Banach space whose dimension is not less than 2, then the complement of each bounded open ball of B is a *continuum. Hence the complements of the *continua of B are a base for the topology of B. If B has dimension 1 and if H is any unbounded proper subset of B that is open and connected, then the complement of H is a *continuum. Hence the complements of the *continua of B are a sub-base for the topology of B. This proves that the collection S contains all Banach spaces. But could the collection S actually include all complete, connected and locally connected metric spaces?

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