Every point in the closure of non-empty set $X$ is either an isolated point or an accumulation point of $\bar{X}$

complex-analysissolution-verification

Suppose $X \subseteq \mathbb{C}$ is non-empty.
We want to show that every point in the closure of $X$ (which is denoted as $\bar{X}$ later) is either an isolated or an accumulation point of $\bar{X}$.

I thought it could be proved by showing contradiction from the negation of the statement, but I think I made a serious mistake that I can't figure out.

Here's the work I've done so far.

Assume that some point $z \in \bar{X}$ is not an isolated point and not an accumulation point of $\bar{X}$.

Since it is not an isolated point of $\bar{X}$, for an arbitrary $\epsilon > 0$, $\exists z_0 \in B(z,\epsilon) \cap \bar{X}$ where $z \neq z_0$.

In other words, $\forall \epsilon > 0, |z-z_0|<\epsilon \cdots \bigstar $ (Note that $z_0 \in \bar{X}$.)

And, since $z_0 \neq z, z_0 \in \bar{X} \setminus \{ z \}$.

Therefore, we can construct a sequence $ \{ z_n \} \subseteq \bar{X} \setminus \{ z \}$ s.t $z_n=z_0$ for any natural number $n$. $\cdots (a)$

Now, replace $z_0$ of equation $\bigstar$ with $z_n$.

Then, we get $\forall \epsilon >0, |z_n – z|<\epsilon$.

And, by $(a)$, we get $\forall \epsilon >0, \exists N \in \mathbb{N}$ s.t $|z_n – z|<\epsilon$ $\ \forall n>N$.

This implies, $z_n \rightarrow z$, thus $z$ is an accumulation point of $\bar{X}$ by definition. (Contradiction!)

Hence, any point in $\bar{X}$ is either an isolated point or an accumulation point. $\blacksquare $

I guess it is wrong. There is not enough reasoning to conclude like this because I didn't use the condition of $\bar {X} $ (A smallest closed set containing $X$), but I can't see how it could be used in this proof. Can I get some help please?

Best Answer

Indeed, you have shown that

For every non-empty subset $Y$ of $\mathbb C$, every points $z\in Y$ is either an isolated point or an accumulation point of $Y$.

I suppose the exercise asks you to show

For every non-empty subset $X$, every $z\in \overline X$ is either isolated or is an accumulation point of $X$ (instead of $\overline X$).

If this is what you want to prove, then after you find $z_n \in \overline X$ so that $z_n \to z$, by definition of $\overline X$, for each $n$, there is $v_n\in X$ so that $|v_n-z_n|<1/n$. Then also $v_n\to z$ and thus $z$ is an accumulation point of $X$.

Remark Concerning your proof, you should be more careful in choosing the sequence $\{z_n\}$: indeed, you choose $\epsilon_n \to 0$ and for each $n$, you find $z_n \in \overline X\setminus\{z\}$ so that $|z_n - z|<\epsilon_n$.

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