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In $\mathbb{R}$, are these sets open? Are they closed?

  1. $A = \{\frac{1}{n} : n \in \mathbb{N}\}$
  2. $B = A \cup \{0\} $
  3. $[0, 1)$

My thoughts:

  1. $A$ is not open as if we have an open ball with $r > 0$ at any point $x$ in $A$ it will contain points that are not in $A$. $A$ is not closed as the complement of $A$ is not open. That is, any open ball at $0$ will contain points from both $A$ and the complement of $A$, namely, 0 is a boundary point.

  2. $B$ is not open as if we have an open ball with $r > 0$ at any point $x$ in $B$ it will contain points that are not in $B$. $B$ is closed as it's complement is a union of open intervals so the complement is open and hence $B$ is closed.

  3. $[0, 1)$ is not open as an open ball with $r > 0$ at 0 will contain points not in the set. It's complement is not open so it is not closed.

How does that look, have I got these correct?

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Looks good as far as I can see, except for what is probably just a typo in (1). I assume what you wanted to say there is "As is not closed as the complement of A is not open. That is, if we have an open ball at 0, it will contains points that are in A, and thus not in the complement of A" – fgp Oct 15 '12 at 23:29
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In 1., at the end, you mean that are in $A$. In 3., maybe you are expected to explain why complement is not open. – André Nicolas Oct 15 '12 at 23:31
    
Here is another post about question 2: math.stackexchange.com/questions/665896/… and – Martin Sleziak Nov 3 '15 at 11:38
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