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I have the following question and i'm not sure how to go about proving whether sets are open closed, or both.

Which of the following sets are open and which are closed?

$1)\ [1,2] \cup [3,4] \ in \ ( \mathbb{R},|\cdot|)$

$2)[1.2] \ in (\mathbb{R},d)$ where d is a discrete metric.

$3)\ B=[x=(x_n)_(n\in\mathbb{N}) :|x_n|< \epsilon\ for \ all \ n \in \mathbb{N}] $ where $\epsilon>0$ is given in $(\ell^\infty, ||\cdot||_\infty)$

$4)\ c_0 $the set of sequences converging to 0 in $(\ell^\infty,||\cdot||~_\infty)$

Any help would be greatly appreciated

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2 Answers


1) Solve this intuitively first. There are basically two types of points in the set (inside an interval or boundary). Is it true that for such points you can always find a little interval that will be contained in the set? Once you convince yourself the answer is no, give a rigorous proof that the set is not open. Now, do the same with the complement and prove the set is closed.

2) Do you know that the union of open sets is open ? (if not, prove it). Do you know that in a discrete space, every singleton set is open (if not, prove it). Now conclude that any set is a discrete space is open (and thus also closed).

3 + 4) Figure out what open balls look like under the supremum metric, then proceed as above.

Remark: 1 and 2 above are very elementary, while 3 and 4 require some more getting used to the abstractness of metric spaces and open sets. Make sure you fully understand 1 and 2 before tackling 3 and 4 (this is the reason why I gave you detailed hints for 1 and 2, and left 3 and 4 more vague).

Good luck!

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For 1) use that union of a finite number of closed sets is closed,

For 2) use that in the discrete metric every set is open and closed.

For 3) use that preimages of open sets are open when the function is continuous

For 4) This one is closed

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The union of closed sets need not be closed (though the union of a finite number does). This can be shown by proving that the intersection of an infinite number of open sets can be closed, for example $$ \bigcap{\infty} (-\frac{1}{n},\frac{1}{n}) = {0} $$. Therefore by taking the complement we get that the union of an infinite number of closed sets need not be closed. –  René G Jan 5 at 5:18
Sorry, that should be $$\bigcap_{\infty} (-\frac{1}{n},\frac{1}{n}) = \left\{ {0} \right\} $$ I made a typo and it won't let me fix it. –  René G Jan 5 at 5:24
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