3
$\begingroup$

In $(\mathbb R^2, ||\cdot||_{\infty})$, let:

$A_0 = ]0,1] \times \{0\}$

$A_n = [(\frac{1}{n}, 0), (\frac{1}{n}, 1)]$ for each $n \ge 1$.

$A = \cup_{n=0}^{\infty} A_n$

It is required to prove that:

$\bar A = A \cup [(0,0), (0,1)]$

I proved that $A \cup [(0,0), (0,1)]$ is contained in $\bar A$.

For the other inclusion, I should probably take a sequence in $A$ and prove that its limit is either in $A$ or in that segment. I did a proof on scratch, but it lacks rigour, and after many attempts, I had no luck constructing a rigorous proof. Hence I am asking for a rigorous proof for that, not a hint, because I already know what is going on.

Though, any other suitable approaches are appreciated.

Thank you.

$\endgroup$
3
  • 1
    $\begingroup$ What does this mean: $ [(0,0), (0,1)]$? $\endgroup$
    – zoli
    Apr 20, 2015 at 21:47
  • $\begingroup$ @zoli, the segment connecting these two points. $\endgroup$
    – user230734
    Apr 20, 2015 at 21:48
  • $\begingroup$ Why you don't just prove that the complement of A∪[(0,0),(0,1)] is open? $\endgroup$
    – san
    Apr 30, 2015 at 17:22

1 Answer 1

1
+50
$\begingroup$

Hint:

So, this is how I picture your sets:

and I added my $B_y$.

For any $y$:

$$B_y=\overline{\bigcup_{n=1}^{\infty}\{(\frac{1}{n},y)\}}={\{(0,y)\}\cup\bigcup_{n=1}^{\infty}\{(\frac{1}{n},y)\}}.$$

Then $$\overline A=\bigcup_{y\in[0,1]}B_y\cup\overline A_0={\bigcup_{n=1}^{\infty}\{(\frac{1}{n},y)\} }\cup A_0\cup \{(0,0)\}\cup [(0,0), (0,1)]=$$= $$={\bigcup_{n=0}^{\infty}A_n }\cup [(0,0), (0,1)]=A\cup[(0,0), (0,1)] .$$

since $$\overline A_0=A_0\cup\{(0,0)\}.$$

As far as the inclusion question see the following cases. $$x=\begin {cases} (\frac{1}{n},y)& \text { for some } n \text { and for some y}\in [0,1],& \text { (case } \alpha)\text{ or}\\ (0,\ y)&\text { for some y}\in ]0,1], &\text { (case } \beta) \text{ or}& \\ (x,\ 0)&\text { for some y}\in [0,1], x\in \ ]0,1] &\text { (case }\gamma)\text{ or}\\ (0,\ 0)&\ &\text { (case } \delta). \end{cases}.$$

$\endgroup$
17
  • $\begingroup$ After second thoughts, I am not quite sure how this proves the other inclusion. I think I misunderstood what went on. Would you mind elaborating a little bit? Thanks. $\endgroup$
    – user230734
    Apr 20, 2015 at 22:43
  • $\begingroup$ You are right. I'll have to go on ... $\endgroup$
    – zoli
    Apr 20, 2015 at 22:55
  • $\begingroup$ Am I right if I think that $]0,1]\times \{0\}=\{x,y: y=0, x\in (0,1]\}?$ $\endgroup$
    – zoli
    Apr 20, 2015 at 22:58
  • $\begingroup$ Sure, that's correct. $\endgroup$
    – user230734
    Apr 20, 2015 at 22:59
  • $\begingroup$ Is this still true: $\overline {]0,1]\times \{0\}}=[0,1]\times \{0\}$ ? $\endgroup$
    – zoli
    Apr 20, 2015 at 23:03

You must log in to answer this question.