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Determine which of the following subsets of $\Bbb R^2$ are open, closed or neither: \begin{align} &1.\quad A=\{(x,y)\in \Bbb R^2:x+y=1\}\\ &2. \quad A=\{(x,y)\in \Bbb R^2: x+y>1\}\\ &3. \quad A=\{(x,y)\in \Bbb R^2:x,y\in \Bbb{Q} \}\\ &4. \quad \Bbb{R}^2\setminus\{1,-1\} \end{align}

  1. closed
  2. open
  3. closed
  4. open

if wrong, please correct me please

I deeply appreciate it

Semiclassical
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chris
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  • For 3 you might consider the definition of open and closed sets and the points made at http://math.stackexchange.com/questions/116721/are-the-rationals-a-closed-or-open-set-in-mathbbr – Henry Sep 26 '14 at 08:01

1 Answers1

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The third set $\{(x,y):x,y\in\mathbb{Q}\}$ is neither open nor closed in $\mathbb{R}^2$, because $\mathbb{Q}$ is neither closed nor open in $\mathbb{R}$. Note that $\mathbb{Q}$ is dense in $\mathbb{R}$, that is every real number is a limit point of $\mathbb{Q}$, however the irrational numbers $\not\in\mathbb{Q}$. Hence $\mathbb{Q}$ is not closed in $\mathbb{R}$. Again take any rational $r\in\mathbb{Q}$. Then any interval $(r-\epsilon,r+\epsilon)\not\subset\mathbb{Q}$ because it contains irrational numbers. Hence $\mathbb{Q}$ is not open in $\mathbb{R}$.

QED
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