In $\mathbb{R}$,

  1. Prove that $A_n= (\frac{-1}{n}, \frac{1}{n})$ is open.

  2. Prove that $A=$ {$a_1, a_2,.....,a_n$} is closed.

  3. Prove that $A_n=$ $[\frac{1}{n}, \infty)$ is closed, and $\cup_n A_n$ is open.

My proofs:

  1. Take $x \in A_n$, and take $\epsilon = \frac{min}{2}$ { $d(\frac{-1}{n}, x), d(x, \frac{1}{n})$}, then $B_{\epsilon} (x) \subset A_n$, is there any need to check this? i.e. to take $y\in B_{\epsilon} (x)$ and prove it's included between $\frac{-1}{n}$ and $\frac{1}{n}$? I feel this is a little hard, is it?

  2. If $\mathbb{R} \setminus A$ is open, then we are done. $\mathbb{R} \setminus A= (-\infty, a_1) \cup (a_1, a_2) \cup ...\cup (a_n, \infty)$ which's a union of open sets, so an open set too.

  3. $\mathbb{R}$ \ $A_n= (-\infty, \frac{1}{n})$ which's an open set, so $A_n$ is closed.

$\cup_n A_n = (0, \infty)$ is open.

  • $\begingroup$ Proofs seem correct to me. $\endgroup$ – Dhruv Kohli - expiTTp1z0 Oct 21 '17 at 18:27
  • $\begingroup$ In 2, you must first state that the numbers are indexed in increasing order. $\endgroup$ – zipirovich Oct 21 '17 at 20:27
  • 1
    $\begingroup$ For 2 note that singletons are closed and A is a finite union of closed sets, henced closed. $\endgroup$ – William Elliot Oct 21 '17 at 21:05

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