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Let $X$ be a topological space.

A subset $U$ of $X$ is sequentially open if each sequence $(x_n)$ in $X$ converging to a point of $U$ is eventually in U. The complement of a sequentially open set is a sequentially closed set, and vice-versa. Every open subset of $X$ is sequentially open and every closed set is sequentially closed. The converses are not generally true.

A sequential space is a space $X$ satisfying one of the following equivalent conditions:

  1. Every sequentially open subset of $X$ is open.
  2. Every sequentially closed subset of $X$ is closed

Why can we say "A sequential space is a space $X$ satisfying one of the following equivalent conditions?

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Oh, come on: each of those statements trivially implies the other. You should be able to prove that in about two lines. –  Brian M. Scott Sep 14 '13 at 10:50

1 Answer 1

A sequentially closed set $C$ is defined as a set containing the limits of all sequences in $C.$

The complement $U$ of a sequentially closed set $C$ is sequentially open: Let $(x_n)_n$ be a sequence converging to $x\in U.$ Then since $C$ is sequentially closed, the sequence $(x_n)$ is not in $C.$ If infinitely many $x_n$ were in $C,$ then they would form a subsequence in $C$ converging to $x,$ thus $x$ would have to be a point in $C.$ Since it isn't, we can follow that $(x_n)_n$ is eventually in $U,$ so $U$ is sequentially open.

The other direction ($U$ sequentially open implies $C$ sequentially closed) is pretty obvious.

Now, the equivalence between the two statements is trivial.

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