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Let $X$ be a topological space and $S=\{x_n\}$ be a sequence of points in $X$. Suppose $a$ is a point in $X$ such that $a$ is adherent to $S$(that is $a$ is in the closure of $S$),I want to ask if there must exist a sequence $\{y_n\}$ in $S$ such that the limit of $\{y_n\}$ is $a$. If not,please give an example,Thanks!

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What do you think? –  Did Jan 18 '13 at 8:14
    
You may want to check out the properties of the Arens-Fort space, which is not a first-countable space. –  Haskell Curry Jan 18 '13 at 8:23
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3 Answers

For another example, let $X$ be the Stone–Čech compactification of $\Bbb{N}$. It does not contain any subspace homeomorphic to $A(\aleph_0)$, i.e., in $\beta \mathbb{N}$ there are no non-trivial convergent sequences (see Engleking's book Corollary 3.6.15).

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The fact that there are no non-trivial convergent sequences in Stone-Čech compactification has been also mentioned (and proven) in Stone-Cech compactifications and limits of sequences. In this blog post proof for $\beta\mathbb N$ is given. –  Martin Sleziak Mar 14 '13 at 6:58
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There is a class of spaces, called Fréchet–Urysohn spaces, which are defined by having the property that $x \in \overline{A}$ iff there is a sequence in $A$ converging to $x$. Agustí Roig's answer indicates that all first-countable spaces are Fréchet–Urysohn. The following example shows that this is a strictly larger class of spaces.

Example 1: Consider the quotient space $X = \mathbb{R} / \mathbb{N}$. (If you are unfamiliar with quotient spaces, let $X = ( \mathbb{R} \setminus \mathbb{N} ) \cup \{ * \}$, and topologise $X$ by declaring $U \subseteq X$ to be open iff $U \setminus \{ * \}$ is open in $\mathbb{R}$, and if $* \in U$, then for each $n \in \mathbb{N}$ there is a $\epsilon > 0$ such that $( n - \epsilon , n ) \cup ( n , n + \epsilon ) \subseteq U$.)

Suppose that $\{ U_n : n \in \mathbb{N} \}$ is any countable family of open neighbourhoods of $*$, for each $n \in \mathbb{N}$ find $\epsilon_n > 0$ such that $( n - \epsilon_n , n ) \cup ( n , n + \epsilon_n ) \subseteq U_n$. Define $$V = \{ * \} \cup \bigcup_{n \in \mathbb{N}} \left( ( n - \frac{\epsilon_n}2 , n ) \cup ( n , n - \frac{\epsilon_n}2 ) \right).$$ Then $V$ is an open neighbourhood of $*$, but $U_n \not\subseteq V$ for all $n$. Therefore $X$ is not first-countable.

To show that $X$ is Fréchet–Urysohn, let $A \subseteq X$, and $x \in \overline{A}$.

  • If $x \neq *$, then letting $A_0 = A \setminus \{ * \}$ it follows that $x \in \mathrm{cl}_\mathbb{R} ( A_0 )$ (the closure of $A_0$ in $\mathbb{R}$ with the usual topology). As $\mathbb{R}$ is Fréchet–Urysohn there is a sequence in $A_0$ converging to $x$ in the topology of $\mathbb{R}$, and this same sequence can be shown to converge to $x$ in the topology of $X$.
  • If $x = *$, the without loss of generality assume that $* \notin A$. I claim that there is a $k \in \mathbb{N}$ such that $k \in \mathrm{cl}_\mathbb{R} ( A )$. If not, then for each $k \in \mathbb{N}$ there is an $\epsilon_k > 0$ such that $( k - \epsilon_k , k + \epsilon_k ) \cap A = \emptyset$ (note that we can take $\epsilon_k < 1$). Then $$V = \{ * \} \cup \bigcup_{k \in \mathbb{N}} \left( ( k - \epsilon_k , k ) \cup ( k , k + \epsilon_k ) \right)$$ is an open neighbourhood of $*$ disjoint from $A$, contradicting that $* \in \overline{A}$! Thus there is a $k \in \mathbb{N}$ such that $k \in \mathrm{cl}_\mathbb{R} ( A )$. As $\mathbb{R}$ is Fréchet–Urysohn there is a sequence in $A$ converging to $k$. This same sequence can be shown to converge to $*$ in the space $X$.

An example of a non-Fréchet–Urysohn space is as follows:

Example 2: Consider $\mathbb{R}$ (or any uncountable set) with the co-countable topology: $U \subseteq \mathbb{R}$ is open iff either $U = \emptyset$ or $\mathbb{R} \setminus U$ is countable. Setting $A = \mathbb{R} \setminus \{ 0 \}$, note that $0 \in \overline{ A }$, however if $( x_i )_{i=1}^\infty$ is any sequence in $A$, then $\mathbb{R} \setminus \{ x_i : i \in \mathbb{N} \}$ is an open neighbourhood of $0$ containing no members of the sequence, and so the sequence cannot converge to $0$.

The above example is somewhat lacking as $X$ fails to be Hausdorff.

Example 3: Consider the ordinal space $X = \omega_1 + 1$ consisting of all ordinals $\leq \omega_1$ (the first uncountable ordinal) with the order topology. This space is easily seen to be Hausdorff, and by properties of ordinals it is also compact. It is easy to see that if $A = \omega_1 = [ 0 , \omega_1 )$, then $\omega_1 \in \overline{A}$. However if $( \xi_n )_{n=1}^\infty$ is any sequence in $A$, then there must be a countable ordinal $\alpha < \omega_1$ such that $\xi_n < \alpha$ for all $n$. Then $( \alpha , \omega_1 ]$ is an open neighbourhood of $\omega_1$ containing no members of the sequence, and so the sequence cannot converge to $\omega_1$.

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Example 3 is good. –  Paul Jan 19 '13 at 1:36
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Generally, for any topological space, the answer is "no". But if the space is a metric (or metrizable) one the answer is "yes". See for instance Munkress' "Topology. A first course", lemma 10.2 (chapter 2.10).

But, in fact, as Munkress warns us, we don't need the full strength of the space being metrizable: it suffices that it satisfies the first countability axiom. See also theorem 1.1 in chapter 4.1: for a space $X$ satisfying the first countability axiom and $A \subset X$, $x\in \overline{A}$ if and only if there is a sequence of points of $A$ converging to $x$.

For the notion of convergent sequence in an arbitrary topological space, see also the definition in chapter 2.10, taking into account that there might be surprises in non-Hausdorff spaces -like sequences converging to more than one point!

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