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I want to show:

Let $(X,\tau)$ be a topological space and $Y\subseteq X$. Then is $\overline{Y}$ closed.

For my proof I want to write $X\setminus\overline{Y}$ as union of open sets.

Proof:

It is $\overline{Y}=Y\cup\partial Y$. Consider $X\setminus(Y\cup\partial Y)=\overline{Y}^c$.

Let $x\in X\setminus(Y\cup\partial Y)$. Hence $x\notin Y$ and $x\notin\partial Y$. Therefor for every neighborhood $U_x$ of $x$ it is $U_x\cap Y=\emptyset$.

Hence $U_x\subseteq X\setminus Y$ and $\overline{Y}^c\subseteq\underbrace{\bigcup_{x\in\overline{Y}^c} U_x}_{\text{open}}\subseteq X\setminus Y$.

Now I want to show, that $\overline{Y}^c\supseteq\bigcup_{x\in\overline{Y}^c} U_x$. Let $x'\in\bigcup_{x\in\overline{Y}^c} U_x$.

I have to show, that $x'\notin Y$ and $x'\notin\partial Y$.

Since $x'\notin\bigcup_{x\in\overline{Y}^c} U_x\subseteq X\setminus Y$ is $x\notin Y$.

Suppose $x'\in\partial Y$. Then holds for every neighborhood $U_{x'}$ of $x'$, that $Y\cap U_{x'}\neq\emptyset$ and $U_{x'}\cap (X\setminus Y)\neq\emptyset$. Which contradicts, that $U_{x'}\subseteq X\setminus Y$.

We condlude, that $\overline{Y}^c=\bigcup_{x\in\overline{Y}^c} U_x$ open. Hence $\overline{Y}$ is closed.

Can you verify my proof? Thanks in advance.

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    $\begingroup$ Reading through this, but the first thing that seems off to me is "Therefor for every neighborhood $U_x$ of $x$ it is $U_x\cap Y=\emptyset$. " - This is not true if we, say, take the neighborhood $U_x = X$ $\endgroup$ – Brevan Ellefsen Jul 14 '18 at 10:08
  • $\begingroup$ Has the neigborhood not to be a subset of $X\setminus Y$? $\endgroup$ – Cornman Jul 14 '18 at 10:10
  • $\begingroup$ I suppose you could be working off a different definition, but a neighborhood of a point $x$ is usually defined to be any open set containing $x$. Another (slightly less common) definition is that a neighborhood is any set containing an open set containing $x$. In either case, the set $X$ is a neighborhood of each of its points $\endgroup$ – Brevan Ellefsen Jul 14 '18 at 10:13
  • $\begingroup$ My definition is, that $U\subseteq X$ is a neighborhood of $x$, if there is an $\epsilon >0$ such that $B(x,\epsilon)\subseteq U$. But you are right. Do you see a way to fix my proof, or does it go down the dumpster? $\endgroup$ – Cornman Jul 14 '18 at 10:17
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    $\begingroup$ Nevermind, we have $\partial Y=\{x\in X| \text{For every neighborhood U of x holds}\,\, U\cap Y\neq\emptyset\,\,\text{and}\,\, U\cap (X\setminus Y)\neq\emptyset\}$. Thats where my conclusion came from. Because $x\notin\partial Y$ this $x$ holds that there exists a neighborhood with $U\cap Y=\emptyset$ or $U\cap (X\setminus Y)=\emptyset$. Also $x\notin Y$. Thats why it has to be $U\cap Y=\emptyset$. Thats at least what I thought. $\endgroup$ – Cornman Jul 14 '18 at 10:24
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I want to show:

Let $(X,\tau)$ be a topological space and $Y\subseteq X$. Then is $\overline{Y}$ closed.

For my proof I want to write $X\setminus\overline{Y}$ as union of open sets.

Sounds like a good approach!

Proof:

It is $\overline{Y}=Y\cup\partial Y$. Consider $X\setminus(Y\cup\partial Y)=\overline{Y}^c$.

Here, I might instead say something like this: "By definition, $\overline Y=Y\cup\partial Y.$ We show that $X\setminus\overline Y$ is a union of open sets, so is open."

Let $x\in X\setminus(Y\cup\partial Y)$. Hence $x\notin Y$ and $x\notin\partial Y$. Therefor for every neighborhood $U_x$ of $x$ it is $U_x\cap Y=\emptyset$.

This isn't quite right. Rather, by definition of $\partial Y,$ since $x\notin Y,$ then there exists a neighborhood $U_x$ of $x$ such that $U_x\cap Y=\emptyset.$

It's also a bit awkward, and there's a misspelling. I might say something like this: "Let $x\in X\setminus\overline Y.$ Since $\overline Y=Y\cup\partial Y,$ then by DeMorgan's Laws, $X\setminus\overline Y=(X\setminus Y)\cap(X\setminus\partial Y),$ so $x\notin Y$ and $x\notin\partial Y.$ Therefore, by definition of $\partial Y,$ there exists some neighborhood $U_x$ of $x$ such that $U_x\cap Y=\emptyset.$ Since $x\in X\setminus\overline Y$ was arbitrary, then such a $U_x$ exists for each such $x.$"

Hence $U_x\subseteq X\setminus Y$ and $\overline{Y}^c\subseteq\underbrace{\bigcup_{x\in\overline{Y}^c} U_x}_{\text{open}}\subseteq X\setminus Y$.

Nicely done! There's really no purpose to mentioning that the union is open right now, though. I'd wait until the end. (See what I do there.)

Now I want to show, that $\overline{Y}^c\supseteq\bigcup_{x\in\overline{Y}^c} U_x$. Let $x'\in\bigcup_{x\in\overline{Y}^c} U_x$.

I have to show, that $x'\notin Y$ and $x'\notin\partial Y$.

You can certainly proceed in this way, though since $X\setminus\overline Y=(X\setminus Y)\cap(X\setminus\partial Y),$ and since you've already shown that $$X\setminus\overline{Y}\subseteq\bigcup_{x\in X\setminus\overline Y}U_x\subseteq X\setminus Y,$$ then you need only show that $$\bigcup_{x\in X\setminus\overline Y}U_x\subseteq X\setminus\partial Y,$$ meaning that you only have to show $x'\notin\partial Y.$

Since $x'\notin\bigcup_{x\in\overline{Y}^c} U_x\subseteq X\setminus Y$ is $x\notin Y$.

This doesn't make sense. It seems like you're trying to say that, since $x'\in\bigcup_{x\in X\setminus\overline Y}U_x\subseteq X\setminus Y,$ then $x'\notin Y.$ However, as I said, we don't even need to say this.

Suppose $x'\in\partial Y$.

That's what I'd do!

Then holds for every neighborhood $U_{x'}$ of $x'$, that $Y\cap U_{x'}\neq\emptyset$ and $U_{x'}\cap (X\setminus Y)\neq\emptyset$. Which contradicts, that $U_{x'}\subseteq X\setminus Y$.

You've got the right idea, but it seems that you're trying to let $U_{x'}$ be simultaneously arbitrary and a specific counterexample. Instead, I'd say something like this: "By definition of $\partial Y,$ this means that for every neighborhood $U$ of $x',$ we have $U\cap Y\neq\emptyset.$ However, since $x'\in\bigcup_{x\in X\setminus\overline Y}U_x,$ then we have that $U_{x'}$ is a neighborhood of $x'$ disjoint from $Y,$ yielding the desired contradiction."

We condlude, that $\overline{Y}^c=\bigcup_{x\in\overline{Y}^c} U_x$ open. Hence $\overline{Y}$ is closed.

Here, I'd just say (if you lead off as I did by announcing your intention) something like: "We conclude that $$X\setminus\overline Y\subseteq\bigcup_{x\in X\setminus\overline Y}U_x\subseteq(X\setminus Y)\cap(X\setminus\partial Y)=X\setminus\overline Y,$$ so that $X\setminus\overline Y=\bigcup_{x\in X\setminus\overline Y}U_x.$ As a union of the open sets $U_x,$ we have that $X\setminus\overline Y$ is open, as we set out to show."


Let me know if you have any questions about my answer, or if you just want to bounce your phrasing adjustments off somebody.

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  • $\begingroup$ Thank you for this answer. I appreciate your comments very much. At the moment I do not have an open question. $\endgroup$ – Cornman Jul 15 '18 at 19:04

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