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Martin Brandenburg
Martin Brandenburg
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Is "second-countable implies separable" equivalent to the Axiom of countable Choice?

It is well-known that a secound-countable topological space is separable. The proof goes like this: Let $(B_n)$ be a (at most) countable base for the topology. We may assume that $B_n$ is nonempty for all $n$, and we may choose some $x_n \in B_n$. Then $D = \{x_n\}$ is easily seen to be dense.

Question This proof needs the Axiom of countable Choice. Can we avoid it? Or is the statement even equivalent to the Axiom of countable Choice? What happens if we also assume that $X$ is a complete metric space?

Is "second-countable implies separable" equivalent to the Axiom of Choice?

It is well-known that a secound-countable topological space is separable. The proof goes like this: Let $(B_n)$ be a (at most) countable base for the topology. We may assume that $B_n$ is nonempty for all $n$, and we may choose some $x_n \in B_n$. Then $D = \{x_n\}$ is easily seen to be dense.

Question This proof needs the Axiom of Choice. Can we avoid it? Or is the statement even equivalent to the Axiom of Choice? What happens if we also assume that $X$ is a complete metric space?

Is "second-countable implies separable" equivalent to the Axiom of countable Choice?

It is well-known that a secound-countable topological space is separable. The proof goes like this: Let $(B_n)$ be a (at most) countable base for the topology. We may assume that $B_n$ is nonempty for all $n$, and we may choose some $x_n \in B_n$. Then $D = \{x_n\}$ is easily seen to be dense.

Question This proof needs the Axiom of countable Choice. Can we avoid it? Or is the statement even equivalent to the Axiom of countable Choice? What happens if we also assume that $X$ is a complete metric space?

Source Link
Martin Brandenburg
Martin Brandenburg
  • 63.1k
  • 13
  • 207
  • 426

Is "second-countable implies separable" equivalent to the Axiom of Choice?

It is well-known that a secound-countable topological space is separable. The proof goes like this: Let $(B_n)$ be a (at most) countable base for the topology. We may assume that $B_n$ is nonempty for all $n$, and we may choose some $x_n \in B_n$. Then $D = \{x_n\}$ is easily seen to be dense.

Question This proof needs the Axiom of Choice. Can we avoid it? Or is the statement even equivalent to the Axiom of Choice? What happens if we also assume that $X$ is a complete metric space?