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  1. Is $B=[0,1]\times (2,3]$ closed in $\Bbb R^2$?
  2. Is ${1}\times \Bbb Z_+$ closed in $\Bbb R^2$?
  3. Is $[0,2]\times [-1,2]$ closed in $\Bbb R^2$?.
  4. Is $A=\{ (x,y) \mid x \ge 0, y \ge 0 \}$ closed in $\Bbb R^2$?

Definition: A set $A$ is closed in $X$ if its complement $X-A$ is open in $X$.

I want to use this definition to show 1., 2. and 3. are closed in $\Bbb R^2$.

For 4. $\Bbb R^2 - A = ( -\infty,0) × \Bbb R \cup \Bbb R\times (-\infty,0)$ is open in $\Bbb R^2$. So $A$ is closed in $\Bbb R^2$. Am I right here?

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    (1). No.! and (3). Yes. – 2017-01-18
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    Yes, you are right about 4. Do yo want to use only open sets of the form $U\times V$ for all parts? – 2017-01-18
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    Yes I need open sets @Hagen von Eitzen – 2017-01-18
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    (2), (3), and (4) are closed in $\mathbb{R}^2$, but (1) isn't. And in the end of your otherwise correct proof for (4), didn't you mean to say that "$A$ is **closed** in $\mathbb{R}^2$", not "*open*"? – 2017-01-18
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    Yes exactly you are right – 2017-01-18

1 Answers 1

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Your reasoning for $(4)$ is correct. You should be able to apply a similar idea to $(3)$, it's just going to be a more complicated expression. You are going to have

$$\Bbb R^2-([0,2]\times[-1,2])=((-\infty,0)\times\Bbb R)\cup(\Bbb R\times(-\infty,-1))\cup C\cup D$$

for some sets $C$ and $D$; you should work out what those are for yourself.

$(2)$ is true, but rather than approaching it the same way as the other ones, try to choose a point $p$ in the complement (call it $V$) and find an open subset $U$ such that $p\in U\subset V$.

For $(1)$, you should be able to find a sequence in $B$ whose limit is not in $B$, showing it isn't closed.

Edit: alternatively for $(1)$, you can write down what the complement $Y$ is, and find a point $p\in Y$ such that there's no open $U$ such that $p\in U\subset Y$, so that $Y$ can't be open.

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    That means I am not able to use given definition to show 1) is open. – 2017-01-18
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    @Kavita: what do you mean? In (1), the set $B$ is neither closed nor open. So you can't show that it's open because it isn't. – 2017-01-18
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    @Kavita see my edit – 2017-01-18
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    Yes I just did mistake to interpret. Yes the set is neither open nor closed. Thanks. – 2017-01-18
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    $$\Bbb R^2-([0,2]\times[-1,2])=((-\infty,0)\times\Bbb R)\cup(\Bbb R\times(-\infty,-1))\cup C\cup D$$ for some sets $C$ and $D$ but I found only one set C =($R$×(2,infinity)). What is D? Need of help. – 2017-01-18
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    @Kavita I'm not doing your homework. I did half of the problem... draw a picture and figure it out; please learn to use this site for its intended purpose, which is **not** a site to cheat on your homework – 2017-01-19
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    I did draw the diagram. But I m not able to find set D. But thanks you did half my work. I just tried to be honest about what I found. Sorry I asked you for more help. – 2017-01-19