My proof of $A^{mathrm{c}}$ is closed iff $A$ is open

Mathematics Asked by mathematicaxyz on January 5, 2022

$$newcommand{c}{mathrm{c}}$$

I have already attempted the proof but I feel like there are some holes in my argument. if someone could look at it and point it out even if minor I would appreciate it. (I know it can be proved through contradiction however I went another way.)

For $$Asubseteq mathbb{R},$$ prove $$A^{mathrm{c}}$$ is closed iff $$A$$ is open.

Suppose $$A^c$$ is closed. Take $$x in A$$ then $$x notin A^c$$. Since a subset $$S$$ in $$mathbb R$$ is closed if $${a_n}$$ in $$S$$ converges to $$a$$, then limit $$a$$ belongs to $$S$$. (Can also say the limit points in $$S$$ are in $$S$$.) Now we know if $$S$$ is closed it contains all of its limit points, then $$x$$ is not a limit point of $$A^c$$. Every neighborhood of $$x$$ such that there is some $$delta > 0$$ such that $$N = { (x – delta,, x + delta)}subset A.$$ Since $$N$$ is the neighborhood of $$x$$ such that $$A^ccap N = emptyset$$, so $$Nsubset A$$. Therefore $$x$$ is an interior point of $$A$$.
$$x$$ is arbitrary so true for any $$x in A$$. Therefore $$A$$ is open if $$A^c$$ is closed.

Now suppose $$A$$ is open (every point in $$A$$ is an interior point). Suppose $$x$$ is a limit point of $$A^c$$. $$x$$ is not an interior point of $$A$$ because it is a limit point of $$A^c$$ and every neighborhood of $$x$$ has a point in $$A^c$$. No point in $$A^c$$ can also be in $$A$$. Therefore $$x in A^c$$. Therefore every limit point of $$A^c$$ is inside of $$A^c$$. From the definition of a closed set $$A^c$$ is closed if $$A$$ is open.

The idea is correct, however, it can be written in a better way. For example, you never write what $$B$$ is. Following is the reworded argument:

First, assume $$A^c$$ is closed. We want to show $$A$$ is open. To that end, let $$x in A.$$ Hence $$x notin A^c.$$ Closedness of $$A^c$$ implies that $$x$$ is not a limit point of $$A^c.$$ Therefore there exists $$delta>0$$ such that $$(x-delta,x+delta) cap A^c=emptyset.$$ Hence $$xin(x-delta,x+delta)subseteq A.$$ This proves that $$x$$ is an interior point of $$A.$$ Since $$xin A$$ was arbitrary, it follows that $$A$$ is open.

Conversely, let $$A$$ be open and we show $$A^c$$ is closed. Suppose $$x$$ is a limit point of $$A^c.$$ We want to prove that $$x in A^c$$. Let if possible, $$x notin A^c.$$ Then $$x in A$$ (open), and so $$x$$ is an interior point of $$A.$$ Therefore there exists a neighborhood $$N$$ of $$x$$ such that $$x in N subseteq A.$$ In other words, $$N cap A^c =emptyset.$$ However as $$x$$ is a limit point of $$A^c,$$ we also must have that $$Ncap A^cneq emptyset.$$ We have thus arrived at a contradiction, which means $$x in A^c.$$ Again since $$x$$ was an arbitrary limit point of $$A^c,$$ the claim follows.

Edit: The converse part can also be proved without using contradiction.

If $$x$$ is a limit point of $$A^c,$$ then for every neighborhood $$N$$ of $$x,$$ there exists $$y_N neq x,$$ such that $$y_N in N cap A^c.$$ In other words, $$y_N in N$$ and $$y_N notin A.$$ This means $$N notsubseteq A.$$ Since $$N$$ was an arbitrary neighbourhood of $$x,$$ this shows that $$x$$ cannot be an interior point of $$A.$$ The openness of $$A$$ thus gives $$x in A^c$$. We have thus shown that $$A^c$$ contains all its limit points and so it is closed.

Answered by Sahiba Arora on January 5, 2022

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