Can someone explain to me why the the subset $B= \{(x,y)|x=0,y\neq 0\} $ in $(\mathbb R^2,d_2)$, is EDIT: not closed? I understand why it's not open. If someone could explain it simply to me that would be great!
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1It's not closed...$(0,0)$ is a limit point but does not belong to $B$. – Viktor Vaughn Nov 27 '16 at 22:49
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Typo sorry! Ah I see, can I ask why is $(0,0)$ a limit point? Is that assuming $x$ and $y$ are approaching 0? – Nov 27 '16 at 22:54
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1Is $d_2$ the usual metric? – Bijesh K.S Nov 27 '16 at 23:02
4 Answers
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You know that this set is closed if R-B is open. The set is open if there is $B(x,r)|r>0 \subseteq R-B$. You can see that if you take the point $(0,0)\in R-B$ there is't a ball that is incluid on it.
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The sequence of points $(0,1/n),$ where $n\in\mathbb{N},$ is contained in $B,$ but this sequence converges to a limit, namely $(0,0),$ and the limit point is not contained in $B.$ Hence $B$ is not closed.
Will R
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Our set is the whole y axis except (0,0). A set is open with a metric if and only if its interior is equal to the set. We can see that the interior of the set is null (there aren't any balls centered at $(o,y), y$ other than $0$, contained in the set). Moreover, a set is closed if and only if it is equal to its closure. It is easy to see by definition, that the closure of our set is the whole y axis. As a result, the set is neither open nor closed.
e_f_i
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Let $\epsilon>0.$ Then $B_\epsilon (0,0)=\{(x,y):\sqrt{x^2+y^2}<\epsilon\}.$ Also for any given $\epsilon>0$ choose $(x,y)\in B$ such that $x=0,y=\epsilon/2.$ Then for any $\epsilon$ we have $\sqrt{0^2+\epsilon^2/4}=\epsilon/2<\epsilon.$
Now you can see that $(0,0)$ is indeed a limit point.
Bijesh K.S
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