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For each of the following metric spaces $(M_i , d_i)$ and subsets $V_i ⊂ M_i$ decide whether $V_i$ is open with respect to the metric topology

1) $M_1= \Bbb R^2,d_1((a,b),(x,y))=|a-x|+|b-y|,V_1=\{(x,y)\in\Bbb R^2|xy>1\}$

2)$M_2=\Bbb R^\Bbb N, d_2(a,b)=$$\sum_{n=1}^{\infty} \frac{2^{-n} |a_n-b_n|}{1+|a_n-b_n|},V_2=\{a \in M_2| \{n \in N \mid a_n \neq 0\} \text{ finite}\}$

Here is my attempt (by the way this is not a homework question its from a past exam I'm using to help me study)

1)if $V$ is in open in $M \implies \exists B_e(a,b) \subset V $ so choosing an arbitrary $(a,b) \in V$ we choose (2,1) as this satisfies $ab>1$ now we have to see if there is an open ball from this point to any other point in $V$. i.e $d((a,b),(x,y))<e$

$d((2,1),(x,y))=|2-x|+|1-y|$ so we need to see is there a finite epsilon s.t. $d<e$ and $x,y$ satisfy $xy<1$

$|2-x|+|1-y|<e$ $\Rightarrow$$\sqrt{(2-x)^2}+\sqrt{(1-y)^2}<e \Rightarrow (2-x)^2+(1-y)^2<e^2=e$

$4-4a+a^2+1-b+b^2<e \Rightarrow a^2-4a-b+b^2<e \Rightarrow a^2(1-\tfrac{4}{a})+b^2(1-\tfrac{1}{b})<e$

and then as we increased the distance between these two points to the furthest apart they can be this implies that $a^2+b^2<e$ and if we suppose these points are still in V this implies that $a^2>\tfrac{1}{b^2}$ and so $a^2+\tfrac{1}{a^2}<a^2+b^2<e \Rightarrow a^2<e \Rightarrow a<e$ and so we have found that epsilon is greater than any real number and so there is no open ball and hence V is not open in M

2) let $a_n=1 \forall n \in \Bbb N $ $d(a,b)=$$\sum_{n=1}^{\infty} \frac{2^{-n} |1-b_n|}{1+|1-b_n|}$ the sum converges to .5 $\Rightarrow$ e>.5 so there is an open ball of radius .5 and so the set is open.

Is this close to right or completely wrong ?, I don't expect it to be totally correct as I'm not very good at this topic yet so any help would be much appreciated.

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For 1)
Suppose $X=\mathbb{R}^2$ and $d(x,y)=|x_1-y_1|+|x_2-y_2|$ and let $V:=\{(x,y)\in X: xy>1\}$.
Let $x=(x_1,x_2)\in V$. Then $\exists\varepsilon>0$ such that $x_1 x_2-1>\varepsilon$. Then $y_1 y_2>1$ for all $y=(y_1,y_2)\in X$ with $d(x,y)<\frac{\varepsilon}{|x_1|+|x_2|}$, since then $$y_1 y_2\geq\left(|x_1|-\frac{\varepsilon}{|x_1|+|x_2|}\right)\left(|x_2|-\frac{\varepsilon}{|x_1|+|x_2|}\right)=|x_1x_2|-\varepsilon+\frac{\varepsilon^2}{(|x_1|+|x_2|)^2}>1$$ Then $V$ is open.
Your mistake was that you did not show it for arbirary $\varepsilon>0$. It is not sufficient to find one $\varepsilon$ such that $y_1y_2<1$ with $d(x,y)<\varepsilon$!
For a contradiction you have to prove this for all $\varepsilon>0$!
I might post something about 2) later but I am a bit in a hurry right now.

edit again: For 2): you cannot just consider $1$ sequence to show that $V$ is open, especially since $a\notin V$.
Let $X=\mathbb{R}^\mathbb{N}$ and $d(x,y)=\sum_{n=1}^\infty\frac{2^{-n}|x_n-y_n|}{1+|x_n-y_n|}$ and $V:=\{x\in X: \#\{x_n\neq 0\}<\infty\}$.
Let $x\in V$ and $I:=\{i\in\mathbb{N}:x_n\neq 0\}$ and suppose $V$ is open. Then $\exists\varepsilon>0$ such that $B_\varepsilon(x)\subseteq V$ (the open ball with radius $\varepsilon$ w.r.t. $d$).
So let $y\in B_\varepsilon(x)$. Then $\exists\delta>0$ such that $\varepsilon-d(x,y)>\delta$.
Then we can find $N\in\mathbb{N}$ such that $\sum_{n=N}^\infty 2^{-n}<\delta$. Then the sequence $(z_n)$ with $z_n=y_n$ for $n<N$ and $z_n=1$ if $n\geq N$ and $y_n=0$. Therefore $V$ is not open.

I hope this helps.

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  • $\begingroup$ Thank you very much , your post was very informative , however I don't see how you got the inequality$ y_1 y_2\geq\left(|x_1|-\frac{\varepsilon}{|x_1|+|x_2|}\right)\left(|x_2|-\frac{\varepsilon}{|x_1|+|x_2|}\right)$ ? $\endgroup$ – excalibirr Jan 9 '18 at 17:52
  • $\begingroup$ sorry my mistake It was supposed to be dependant on $M_2$ I've edited the question to reflect that now :) $\endgroup$ – excalibirr Jan 9 '18 at 17:54
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    $\begingroup$ First we see that $x_1x_2>0$ for all $(x_1,x_2)\in V$. First suppose $x_1,x_2>0$. Let $\delta:=\frac{\varepsilon}{x_1+x_2}$. If we suppose that $d(x,y)<\delta$ then $|x_1-y_1|<\delta$ and also $|x_2-y_2|<\delta$. Then the inequality follows. If $x_1,x_2<0$ then we have a similar argument, since then $y_1,y_2<0$, because always at least $|x_1|>1$ or $|x_2|>1$. feel free to ask further if something is not clear. $\endgroup$ – Pink Panther Jan 9 '18 at 18:02
  • $\begingroup$ It took me a while but I see how the inequality was obtained now ( my brain is slow from 10 hours straight study haha). when you decided to choose $d(x,y) < \tfrac{e}{|x_1|+|x_2|} $ did you know you would get this inequality or what thoughts went into choosing it ? $\endgroup$ – excalibirr Jan 9 '18 at 18:23
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    $\begingroup$ Well I pretty much started with guessing that bound for $d(x,y)$. I first i tried with $\frac{\varepsilon}{4}$, which did not work apparently. Then i started to look at that product in the middle term and looked for a bound so that $|x_1|+|x_2|$ cancels with the bound so that the whole term becomes greater than $1$. $\endgroup$ – Pink Panther Jan 9 '18 at 18:28

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