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This question is a little bit easy but I would appreciate if you can help me.

I am familiar with definition and theorems about closed spaces. However, when I see a set I cannot determine that it is closed or not.

In some answers people say that "this set is clearly closed". But it is not clear to me :)

For instance $A = \{(t,0): t> 0\}$ and $B=\left\{\left(t,\dfrac1t\right): t>0\right\}$. What should I look at to understand they are closed or not? (without a long proof if possible)

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I guess you mean these are closed in $\mathbb{R^2}$. –  Erno Nemecsek Dec 10 '11 at 13:11
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For the set $B$, when $t=0$, $\frac{1}{t}$ is not defined. –  Paul Dec 10 '11 at 13:12
    
@Ozan Yeah sorry, I forgot :) Of course $\mathbb{R}^2$. I saw those examples on a site and i did not realize t equals to $0$. What if we say: A={(t,0):t>0} and B={(t,1t):t>0}. Are they closed? –  marvinthemartian Dec 10 '11 at 13:16
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3 Answers

up vote 2 down vote accepted

For the set $$B=\left\{\left(t,\frac1t\right): t>0\right\}$$ you argue that as $x$ goes to $\infty$, $1/x$ goes to $0$, but $0\notin B$, so $B$ isn’t closed. The problem with this argument is that $B$ isn’t a set of real numbers: it’s a set of ordered pairs of real numbers. The question isn’t whether $0\in B$ or not: we know that $0\not \in B$, because it isn’t even the kind of thing that can belong to $B$. The question is whether there is some point $(a,b)\notin B$ such that every open nbhd of $(a,b)$ contains a point of $B$.

Suppose, then, that the point $p=(a,b)\notin B$. If you make a rough sketch of $B$, you shouldn’t have any trouble convincing yourself that if $p$ is below and to the left of $B$, there are real numbers $c>a$ and $d>b$ such that $$\{(x,y):x<c\text{ and }y<d\}$$ is an open nbhd of $p$ disjoint from $B$: it’s a lower lefthand ‘quadrant’ of the plane lying entirely below and to the left of $B$.

Similarly, if $p$ is above and to the right of $B$, there are real numbers $c<a$ and $d<b$ such that $$\{(x,y):x>c\text{ and }y>d\}$$ is an open nbhd of $p$ disjoint from $B$: it’s an upper righthand ‘quadrant’ of the plane lying entirely above and to the right of $B$.

Thus, every point of $\mathbb{R}^2\setminus B$ has an open nbhd disjoint from $B$, and $B$ is therefore closed.

If you actually had to prove that $B$ is closed, you’d have to explain how to find $c$ and $d$ for different values of $a$ and $b$, but if you’re simply trying to decide for yourself whether $B$ is closed, this sort of pictorial reasoning is just fine.

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In metric spaces, is a very good exercise to you to prove the following:

Theorem. Let $C$ a subset of a metric space $E$. The following statements are equivalent:

  • $C$ is closed.
  • For every sequence $(x_k)$ in $C$ if $x_k\to x$, for some $x\in E$, then $x\in C$.

For example for your set $A$, take the sequence $\{(1/k,0)\}$. Clearly $(1/k,0)\in A$ for every $k\in\mathbb{N}$, but $(1/k,0)\to (0,0)\not \in A$, therefore $A$ isn't closed.

I think that is easiest way to see that a set is closed or not, at least in metric spaces.

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We can think of the complement of $A$, let's denote it by $A^c$ which is the set $\mathbb{R^2} - A$. Take any point $p$ in the complement. Then can you find an open ball around this point such that the open ball is contained in the complement? If you cannot see, try drawing a picture, it helps a lot.


Hint: The point $(0,0)$ is in the complement of $A$. Can you find an open ball around it?

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Thank you for oyur answer! (Sağol ;)) What confuses me about the set B is as x goes to infinity 1/x goes to zero. But $0$ is not in the set B. Then B is not closed since it has an accumulation point which is not in B. What is wrong with my logic? –  marvinthemartian Dec 10 '11 at 13:36
    
You're welcome! (Rica ederim :)) Your logic is true actually but sometimes it is easier to decide the set is closed or not by looking at its complement. I think in most cases it is hard to determine the accumulation points of the sets. –  Erno Nemecsek Dec 10 '11 at 14:08
    
The 2 methods are essentially the same: A point $p$ in the complement of $B$ is an accumulation point of $B$ iff every neighborhood of $p$ intersects $B$. –  Ted Dec 10 '11 at 18:48
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