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Let $(X, \tau)$ be a topological space, $A \subset X, x \in A$.

The lecture notes that I'm working with define $x$ as an accumulation point of $A$ iff there exists a sequence $\{x_n \} \subset A \setminus \{x\}$ for which $x_n \to x$ (I'm aware that this may not be the canonical definition, e.g. there's a different definition here) and define $x$ as an isolated point of $A$ if there is an open set that contains $x$ and no other points.

The notes claim that every point in $A$ is either isolated or an accumulation point.

I am having trouble proving this in a general space (in fact I'm not even sure that it's true). Clearly isolated $\implies$ not accumulation. On the other hand suppose a point $x$ is not isolated, then we need to find a sequence $\{x_n\}$ with $x_n \to x$. I would like to say something like "let $\{U_n \mid n \in \mathbb N\}$ be a sequence of open sets containing $x$, then $\left\{\bigcap_{k=1}^n U_k \mid n \in \mathbb N\right\}$ is also a sequence of open sets, so choose an $x_n \in \bigcap_{k=1}^n U_k$ for each $n$ to form the sequence". But the definition of $x_n \to x$ involves all open sets containing $x$, of which there could be uncountably infinitely many, so this approach does not seem to work.

D G
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    If "sequence" means "sequence indexed by $\mathbb N$", this is indeed false. You can take the ordinal $\omega_1 + 1$ with the order topology. Then $\omega_1$ is not isolated, but it is not the limit of a sequence of lower ordinals (because its cofinality is $\omega_1$). – Mees de Vries Mar 13 '20 at 11:49
  • The def'n of a topology is very broad. There are topologies that cannot be described by countable sequences. – DanielWainfleet Mar 13 '20 at 21:14
  • @DG. Either assert that $X$ is second countable or allow the canonical definition ; i.e. define $x$ as an $\it{accumulation};\it{point}$ of $A$ if there exists a net in $A-{x}$ converging to $x$. Equivalently, $x$ is an accumulation point of $A$ when $x\in cl(A-{x})$. – Oliver Kayende Mar 14 '20 at 00:47

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This is false. If $x$ is not isolated point of $A$ you can only say that there is net in $A$ converging to $x$. A sequence converging to $x$ need not exist.

For example consider $[0,1]^{[0,1]}$ with the product topology and let $ A=C[0,1]$. Let $f$ be a function which is discontinuous at every point. Then $f$ is a limit point of $A$ but there is no sequence from $A$ converging to $f$. This follows from the fact that the pointwise limit of a sequence of continuous functions is continuous on a dense set.

Kavi Rama Murthy
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Let $X$ be an uncountable set with the co-countable topology: Any $Y\subset X$ is open iff $[Y=\phi$ or $X \setminus Y$ is countable]. (Countable means "finite or countably infinite".)

There are no non-empty finite open sets, hence no isolated points.

But when $x\in X$ and $(x_n)_{n\in \Bbb N}$ is a sequence in $X$ \ $\{x\}$ then $X$ \ $\{x_n:n\in \Bbb N\}$ is a nbhd of $x$ that contains $no$ member of the sequence. So we do not get $x_n\to x.$

DanielWainfleet
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