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I guess one way to explain that $\mathbb{R}$ is open:

  • because $\emptyset$ is closed,

    since $\mathbb{R} = \mathbb{R} \backslash \emptyset$,

    then $\mathbb{R}$ is open. <$\because$ the definition of closed sets>

However,

  • I am thinking $\mathbb{R}$ = $S \cup \mathbb{R} \backslash S$, where $S \subset \mathbb{R}$,

    and $S_1 \cup S_2$ is open if $S_1, S_2$ are open. <$\because$ the union of a collection of open sets is open>

    Let $S_1$ be any $S \subset \mathbb{R}$,

    Let $S_2$ be $\mathbb{R} \backslash S$,

    Then $S_1 \cup S_2$ is not open, since $\mathbb{R} \backslash S$ is closed.

    Therefore, $\mathbb{R}$ is not open.

If there is something wrong with my logic, can someone please point them out for me?, Thank you

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closed as unclear what you're asking by Shaun, TheGeekGreek, Namaste, B. Goddard, user223391 Sep 8 '17 at 18:07

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ – Shaun Sep 4 '17 at 14:01
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    $\begingroup$ Welcome to Math.SE,is my edit okay? $\endgroup$ – kingW3 Sep 4 '17 at 14:05
  • $\begingroup$ Your logic fails at "therefore $\mathbb{R}$ is not open". No, what you've shown is that $\mathbb{R}$ is closed. A topological space $X$ can have subsets which are closed and open at the same time, with a notable example of $\emptyset$ and $X$ itself. Read this for more details: en.wikipedia.org/wiki/Clopen_set $\endgroup$ – freakish Sep 4 '17 at 14:21
  • $\begingroup$ @freakish: I disagree that the argument shows anything. If $S$ had been specified to be a clopen subset, then the argument readily shows that $\Bbb R$ is closed. Hwever, $S$ was arbitrary, so nothing of a topological nature was shown about $\Bbb R.$ $\endgroup$ – Cameron Buie Sep 4 '17 at 14:30
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It is true that if two sets are open, then their union is open. However, the converse need not hold. For example, $$(0,1)=\bigl[(0,1)\cap\Bbb Q\bigr]\cup\bigl[(0,1)\setminus\Bbb Q\bigr],$$ so an open set can be a union of two non-open sets. Also, $$(0,\infty)=(0,1)\cup[1,\infty),$$ so an open set may be a union of an open set and a closed set.

Another thing to keep in mind is that sets can be open, closed, both, or neither. The two sets in the first union are neither, while their union is open. Also, you haven't justified that $\Bbb R\setminus S$ is closed, since you never specified that $S$ was open.

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Sets are not doors.

$\Bbb R$ is both open and closed in $\Bbb R$, as is every set in itself.

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    $\begingroup$ I beg you pardon, but painter Marcel Duchamp in the 1920's realised in his apartment a door that was both open and closed. $\endgroup$ – Bernard Sep 4 '17 at 14:12
  • $\begingroup$ This does not appear to answer the question, despite multiple upvotes. The OP never questioned whether $\Bbb R$ was open, and never addressed whether or not $\Bbb R$ was closed. Rather, the OP asked what went wrong in the given reasoning. $\endgroup$ – Cameron Buie Sep 4 '17 at 16:45

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