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Is it correct to say:

If a set $A$ has a point $x$ such that for all $r>0$, the open ball of radius $r$, centered at $x$ is not a subset of $A$, then $A$ must be a closed set.

Andrés E. Caicedo
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amazonprime
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    Unfortunately not, this only implies $A$ is not open. – ocg Jan 26 '16 at 04:34
  • @JulienGodawatta

    Interesting. So in real analysis, "not open" is not the same as "closed"? I thought 'open' and 'closed' were negations of each other :(

    Does this mean that there can be a set that is not open nor closed?

     – amazonprime Jan 26 '16 at 04:39
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    Indeed, unfortunate terminology. Sets can be open, closed, neither, or both ("clopen"). – ocg Jan 26 '16 at 04:40
  • @JulienGodawatta

    Hmm. Good thing I didn't end up relying on this in my homework. Thank you for the info!

     – amazonprime Jan 26 '16 at 04:43
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    @JulienGodawatta It may be useful for you to know that the complement of a closed set is open. – Adam Francey Jan 26 '16 at 04:49
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    Some familiar examples on the real line: $(0,1)$ is open, $[0,1]$ is closed, $[0,1)$ is neither open nor closed, and $\mathbb{R}$ is both open and closed. – Austin Mohr Jan 26 '16 at 05:00
  • The condition says that $x$ is a limit point of $X\setminus A$, the complement of $A$. This implies that the complement is not closed, otherwise $x$ would be in that set whereas it's in $A$. Therefore $A$ is not open. But you can't conclude that $A$ is closed: Re your question above — correct: "not open" $\ne$ "closed". What is true is that the complement of an open set is closed, and the complement of a closed set is open: $A$ is closed [resp. open] iff $X\setminus A$ is open [resp. closed]. – BrianO Jan 26 '16 at 05:05

1 Answers1

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As has been said in the comments, your formulation is not quite correct.

Here is something that you can say:

$A$ is closed iff, for each $x \not \in A$, there is a $r > 0$ so that $B_r(x)$ does not intersect $A$.

Hint for proof: Try to show that $A^c$ is open. Show that the complement of an open set is closed. (And conversely, the complement of an open set is closed.)

Further relevant exercises, related to the comments: 1) Show that $\mathbb{R}$ and $\emptyset$ are both both open and closed in $\mathbb{R}$. Generalize this to a general metric space. 2) Find a metric space in which every subset is both closed and open. 3) Describe the open and closed sets in $\mathbb{Q}$ 4) Describe the closed and open sets in the Cantor set.

Elle Najt
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