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Hi¡ I have a problem with the next exercise.

Let $(X,\tau)$ be a topological space. Prove that $\mathcal{U}\subset X$ is open in $X$ if and only if for all $\mathcal{A}\subset X$ holds that $\text{cl}(\mathcal{U}\cap\text{cl}(\mathcal{A}))=\text{cl}(\mathcal{U}\cap \mathcal{A})$ where $\text{cl}()$ is the closure of a set and $\text{der}()$ is the derived set.

My attempt:

$\Rightarrow)$

Let $\mathcal{A}\subset X$

$\supseteq)$

Then, $\mathcal{U}\cap\mathcal{A}\subseteq\mathcal{U}\cap\text{cl}(\mathcal{A})$ and therefore, $\text{cl}(\mathcal{U}\cap\mathcal{A})\subseteq\text{cl}(\mathcal{U}\cap\text{cl}(\mathcal{A}))$

For the another contention, the same:

$\subseteq)$

Let $x\in\text{cl}(\mathcal{U}\cap\text{cl}(\mathcal{A}))$, then, $x\in\mathcal{U}\cap\text{cl}(\mathcal{A)}$ or $x\in\text{der}(\mathcal{U}\cap\text{cl}(\mathcal{A}))$

If $x\in\mathcal{U}\cap\text{cl}(\mathcal{A})$, then, $x\in \mathcal{U}$ and ($x\in \mathcal{A}$ or $x\in\text{der}{\mathcal{A}}$)

If $x\in \mathcal{A}$ and $x\in \mathcal{U}$, then, clearly, $x\in\text{cl}(\mathcal{U}\cap\mathcal{A})$

If $x\in\mathcal{U}$ and $x\in\text{der}(\mathcal{A})$, then, $x\in\text{cl}(\mathcal{U})\cap\text{cl}(\mathcal{A})$, but, here, again, I'm very stuck.

For the $\Leftarrow)$ I don´t know how can I do it. I need a hint, or more.

Thanks in advance.

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I think it's best to use the characterisation

$x \in \overline{A}$ iff for all open sets $O$ that contain $x$, $O \cap A \neq \emptyset$, instead of the formula $\overline{A} = A \cup A'$ (where the latter is the derived set of $A$, its limit points), and doing a case analysis.

I wrote up a complete proof here, using this characterisation.

Inspired by the erroneous proof using sequences by inquisitive, I came up with a proof using nets; I will not give it (except upon request), because most courses do not treat nets at all.

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continuing your proof you can say that,

since $x \in \overline{\mathcal{A}}$, $\exists$ a sequence $\{x_i\} \in \mathcal{A},\ \{x_i\} \rightarrow x$,

and since $x \in \mathcal{U}$ and $\mathcal{U}$ is open we can find a $N \in \mathbb{N}$ st $\forall n > N, x_n \in \mathcal{U}$, thus we have a sequence $\{x_i , i>n \} \in \mathcal{A} \cap \mathcal{U} $ converging to $x$.

thus, $x \in \overline{\mathcal{A} \cap \mathcal{U}} $

for the reverse you can prove if $\overline{\mathcal{A} \cap \mathcal{U}} = \mathcal{A} \cap \overline{\mathcal{U}} \ \ \forall \ \mathcal{A} \subset X$ then, $ X \setminus \mathcal{U}$ is closed.

to prove this, let $y \in der(X \setminus \mathcal{U})$ s.t $y\notin X \setminus \mathcal{U}$

then $\exists$ a sequence $\{y_i\} \in X \setminus \mathcal{U}$ converging to $y$

let $\mathcal{A} = \{y_i , i\in \mathbb{N}\}$

now we know that $y\in \overline{\mathcal{A}}$

Since $\mathcal{A} \subset X \setminus \mathcal{U}$, therefore, $\mathcal{U}\cap\mathcal{A} = \overline{\mathcal{U}\cap\mathcal{A}} = \varnothing$

But, $y \in \mathcal{U} \cap \overline{\mathcal{A}}$, which is a contradiction.

Hence, $y \in X \setminus \mathcal{U} \implies \mathcal{U}$ is open.

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  • $\begingroup$ There does not need to be such a sequence ,so this is not a general proof. $\endgroup$ – Henno Brandsma Mar 2 '17 at 19:54

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