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In a general topological space are these properties equivalent ? If not, is there a property (e.g. first countability) that metric spaces possess which makes them equivalent there ? Here are the definitions as I understand them.

Bolzano-Weierstrass Property: every infinite sequence $(x_n)$ in X has an accumulation point in X, i.e. a point x such that every open set that contains x contains an infinite number of the points in $(x_n)$

Sequentially compactness: every infinite sequence in X has a convergent subsequence, i.e. converges to a limit, i.e. for every open set O in the topology there is N such that for all n > N then $x_n$ ∊ O (where $(x_n)$ are the points in the subsequence).

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The two properties are equivalent if and only if the topological space admits local bases countable, i.e. $(X, \mathcal{T})$ is $N_1$.

In fact, if X is a space sequentially compact, then for all $\lbrace x_n \rbrace \subset X$ exists $x_{n_k} \rightarrow x$, and it is easy to conclude that $x$ is a point of accumulation for $\lbrace x_n \rbrace \subset X$. Conversely, if $x$ is a point of accumulation for $\lbrace x_n \rbrace \subset X$, considers a countable local base $\mathcal{U}(x)=\lbrace U_m : m \in \mathbb{N} \rbrace$. We can assume that $U_{m+1} \subset U_m$ $\forall m \in \mathbb{N}$ (why?). Since $x$ is an accumulation point, we can define a sequence of integers $m_k$ such that $m_{k} < m_{k+1}$ with $x_{m_k} \in U_{m}$, consequently $x_{m_k} \rightarrow x$.

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  • $\begingroup$ I think it has to be $T_1$ as well as first countable ? $\endgroup$ – Tom Collinge Dec 20 '15 at 19:26
  • $\begingroup$ With $N_1$ I denote that the topological space admits local bases countable (or neighborhood bases countable). In this case I would say that $T_1=N_1$. $\endgroup$ – user288972 Dec 20 '15 at 19:35

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