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Let $X =\mathbb{N}\times \mathbb{Q}$ with the subspace topology of $\mathbb{R}^2$ and $P = \{(n, \frac{1}{n}): n\in \mathbb{N}\}$ .

Then in the space $X$

Pick out the true statements

1 $P$ is closed but not open

2 $P$ is open but not closed

3 $P$ is both open and closed

4 $P$ is neither open nor closed what can we say about boundary of $P$ in $X$?

I always struggle to figure out subspace topology. Though i am aware of basic definition and theory of subspace topology. I need a bit explanation here about how to find out subspace topology of $P$?

Thanks for care

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It is the subspace topology of $X$, not $P$. $P$ is a subset of $X$. Open sets in $X$ are open sets of $\mathbb{R}^2$ intersected with $X$. –  copper.hat May 13 '12 at 19:28
    
Oh thank you sir –  srijan May 13 '12 at 19:36
    
This is not the first time you use "Pick out the true statements" in a title, try to avoid such title. Instead try to make the titles informative so people will have a clue what is in the question without having to open it. –  Asaf Karagila May 13 '12 at 19:51
    
@AsafKaragila Surely i will take care of that. –  srijan May 13 '12 at 19:54
    
Long time back I asked these questions where I used to use "pick out .." At the time I was new to this website. But today I am getting negative votes without mentioning any reason. It is something heartening me. –  srijan Sep 18 '12 at 5:03
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2 Answers 2

up vote 3 down vote accepted

Draw a picture: $P$ is a set of points on the positive half of the graph of the hyperbola $y=\frac1x$. Let’s look at one of those points, say $\left\langle 4,\frac14\right\rangle$. Is $\{4\}$ an open set in $\Bbb N$? Is $\Bbb Q$ an open set in $\Bbb Q$? If the answers to these questions are both yes, then $$\{4\}\times\Bbb Q=\{\langle 4,y\rangle:y\in\Bbb Q\}$$ is an open set in $X$. Call this set $U$; what is $U\cap P$?

This should help in deciding whether $P$ is open in $X$. To decide whether $P$ is closed in $X$, you need to consider a point $\langle n,q\rangle\in X\setminus P$ and ask whether this point can possibly be a limit point of $P$. It will help to realize that sets of the form $\{n\}\times(q-\epsilon,q+\epsilon)$ are open in $P$, because $\{n\}$ is open in $\Bbb N$, and $(q-\epsilon,q+\epsilon)$ is open in $\Bbb Q$. (Here I’m taking the interval in $\Bbb Q$, not in $\Bbb R$.)

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+1 : ) ${}{}{}$ –  Matt N. May 13 '12 at 19:32
    
@Brian Really wonderful way of explaining the things.Sir, Can you tell me about $U\cap P$? What should be approach? –  srijan May 13 '12 at 19:41
    
@srijan: Suppose that $\langle x,y\rangle\in U\cap P$; then $x=4$; what’s the only point of $P$ that has its first coordinate equal to $4$? (Actually, my $U$ is more complicated than necessary, and I’m about to change it to $\{4\}\times\Bbb Q$. Everything that I’ve said in this comment will still apply.) –  Brian M. Scott May 13 '12 at 19:50
    
@BrianM.Scott Now its clear to me. I am happy. Thank you sir:) –  srijan May 13 '12 at 19:56
1  
@srijan: No, the boundary of $P$ will be $P$ itself. Remember, a point is in the boundary of $P$ iff it is in $\operatorname{cl}P\cap\operatorname{cl}(X\setminus P)=P\cap\operatorname{cl}(X\setminus P)$, since $P$ is closed. But every point of $P\subseteq\operatorname{cl}(X\setminus P)$, so this intersection is just $P$ itself. –  Brian M. Scott May 13 '12 at 20:05
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For example, 1 is true. You can see that $P$ is not open by looking at an $\varepsilon$-ball around any point $p = (n, \frac1n )$ in $P$. Then there will be a rational $q$ such that $(n,q)$ is inside the ball hence $P$ is not open. (because $\mathbb Q$ is dense in $\mathbb R$)

Also, it's closed: think about why its complement is open. (You can make an $\varepsilon$-ball around the point that doesn't intersect with $P$)

Now that we have established that 1 is true, we know that 2, 3 and 4 are false.

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@ Matt Thank you sir. I got to know about sophisticated approach to tackle this kind of problems. –  srijan May 13 '12 at 19:45
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