An alternative definition of open sets using the following property?

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Recently I have gone through the following problem.

Let $E$ be a subset of a metric space with the following property: If a sequence $(x_n)_{n\in\mathbb{N}}\in X$ converges to $x\in E$ then there exists $n_0\in\mathbb{N}$ such that $x_n\in E$ for all $n\ge n_0$. Prove that $E$ is open.

Before proving this, I have proved that,

Let $(X,\tau)$ be a topological space and $Y\subseteq X$. Then $Y$ is closed iff it contains all its adherent points.

After writing a complete proof I found that if we assume that in a topological space $(X,\tau)$ every adherent point of $Y\subseteq X$ is the limit of some sequence $(x_n)_{n\in\mathbb{N}}\in Y$ then (using the above theorem) we can give a definition of open sets of that topological space in terms of sequences in the following way,

Let $X$ be a non-emptyset and $E\subseteq X$. We will say $E$ to be an open set in $X$ if a sequence $(x_n)_{n\in\mathbb{N}}\in X$ converges to $x\in E$ implies the existence of $n_0\in\mathbb{N}$ such that $x_n\in E$ for all $n\ge n_0$.

My questions are,

  • Can the above idea be used as an alternative to define open sets of arbitrary topological spaces, not necessarily in terms of sequences but maybe for some suitable generalization of the notion of sequences?

  • If so, then can some related literature be mentioned?

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As Tim Raczkowski says, your definition does not work as is (consider the uncountable well ordered set with a point at $\infty$). But there is a generalization of sequence which may help. That generalization is nets. Nets are mappings of arbitrary directed sets into the topological space, so they are like sequences, but may have uncountably many terms. For more information, you could take a look at General Topology by Stephen Willard. You can even see the original exposition of nets here, though the article is almost 100 years old, and a bit dense.

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This idea only works for topological spaces which are 1-st countable. That is, that every point has a countable basis. We can't define closed spaces in terms of sequences because of this. If a space $X$ is not 1-st countable, it may have too many open sets to guarantee that a closed set contains all it's sequential limits.

Check Munkres - Topology $\S 21$.