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How to prove this please

" Let $(E,\theta)$ a Hausdorff topological space and $K$ compact, let $A\subset K$ such that $card(A)=+\infty$, then $A$ has at least an accumulation point"

We say that $x$ is an accumulation point of $A$ iff $\forall V\in \mathcal{V}_{x}, (V\setminus\{x\})\cap A\neq \emptyset$

But i don't know how to do the proof .

Thank you

2 Answers 2

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Assume $A$ has no accumulation point. Then every $x\in K$ has an open neighborhood $U_x$ with $(U_x\setminus \{x\})\cap A=\varnothing$ (i.e. $x$ is no accumulation point). These $U_x$ are an open cover of $K$ and because $K$ is compact, only finitely many suffice to cover all of $K$, say $U_1,...,U_n$. The $U_i$ also cover all of $A$ as $A\subseteq K$. So for each $x\in A$ there must be a $U_i$ with $x\in U_i$. As $A$ has infinitely many points, but there are only finitely many $U_i$, one of the $U_i$ must contain infinitely many points of $A$, i.e. $U_i\cap A=\{\mathrm{infinite~set}\}$. This is in contradiction to the definition of the $U_x$.

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    i don't understand: Each $U_i$ can contain at most one point of $A$ ? – 2017-02-18
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    Look at how the $U_x$ are defined! They are defined to contain _no_ point of $A$ with the possible exception of the "center point". This center point can be from $A$, but no other point. – 2017-02-18
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    yes. Every point in $A$ is isolated with respect to the subspace topology, which is equivalent to what he wrote in the second sentence. – 2017-02-18
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    Who is the "center point " ? – 2017-02-18
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    and why: Each $U_i$ can contain at most one point of A. But A is infinite and there are only finitely many $U_i$ – 2017-02-18
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    The $x$ is the "center point" of $U_x$. I will try to elaborate. – 2017-02-18
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    the $x\in A$ is not the same $x\in K$ no ? – 2017-02-18
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    No. You can see that I talk about _every_ $x\in K$ and _each_ $x\in A$ respectively. – 2017-02-18
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You can even show more. Suppose $A$ is infinite. Then there exists $x \in X$ such that

$$\forall U \in \mathscr{V}_x: U \cap A \text{ is infinite}$$

Such an $x$ is called an accmulation point of the set $A$,and it's certainly an accumulation point in your sense as well (although your definition is usually called a limit point of $A$). (See the final paragraph as well)

In fact if you covered cardinalities (denoted by $||$) and know some set theory, you can also show the generally stronger statement that there exists $x \in X$ such that

$$\forall U \in \mathscr{V}_x: \left|U \cap A\right| = \left|A\right|$$

which is called a point of total accumulation. As $A$ is infinite this is at least as strong as the previous notion of accumulation point, but stronger for uncountable $A$, in general.

To prove this final statement: suppose no such $x$ exists, this means by logical negation of the definition:

$$\forall x\in X: \exists U_x \in \mathscr{V}_x: |A \cap U| < |A|$$

(As $A\cap U \subset A$, it's cardinaility is either smaller or equal to that o $A$, and being a non-"point of total accumulation" gives us at least one where it's smaller).

Then $\{U_x: x \in X\}$ is by definition an open cover of $X$ (each $x$ covered at least by its "own" $U_x$) so there is a finite subset $I \subseteq X$ such that $X = \cup\{U_x: x \in I\}$ by compactness.

But then $$|A| = A \cap X| = |A \cap (\cup\{U_x: x \in I\})| = |\cup\{U_x \cap A: x \in I\}| \le \sum_{i \in I} |A \cap U_i| < |A|$$

The last part uses that for infinite cardinalities, a finite sum is just the maximum, etc. This final contradiction shows us that at least one such $x$ must exist.

And this is certainly an accumulation point in your sense, because for any $V \in \mathscr{V}_x$, $(V \cap A) \setminus \{x\}$ is non-empty, even infinite, as we remove a single point from an intersection $A \cap U$ which has the same size as $A$, so is infinite. The basic trivial observation is that

$$(V\setminus \{x\})\cap A = V \cap (A \setminus \{x\} = (V \cap A) \setminus \{x\}$$

Possibly interesting final fact: $X$ is compact iff every infinite subset $A$ of $X$ has a point of complete accumulation. This is the way Aleksandrov and Urysohn proved Tikhonov's theorem, I believe.