1
$\begingroup$

I want to find the closure and interior of $[0,1)$ in $\mathbb{C}$. Intuitively, the interior is empty and the closure is just $[0,1]$ just like its closure in $\mathbb{R}$. I am having trouble formally proving the latter from the definition. I guess the problem boils down to showing that sequences in $[0,1)$ only converge to values in $[0,1]$? This is intuitive, but I'm just not sure how to say it rigorously.

$\endgroup$
  • $\begingroup$ Identify $\mathbf{C}$ as $\mathbf{R}^2$. $\endgroup$ – IAmNoOne Nov 7 '16 at 7:49
  • $\begingroup$ @Nameless was that a suggestion for me to edit the question or a hint? If the latter, I am already kind of thinking that way $\endgroup$ – b_pcakes Nov 7 '16 at 7:51
  • 1
    $\begingroup$ @b_pcakes Its a hint, you should think about closed sets in the product topology. $\endgroup$ – user259242 Nov 7 '16 at 7:56
  • $\begingroup$ Can you please expand a little? I want to show that $\overline{[0, 1)} \subset [0,1]$, but how do I do that by definition? $\endgroup$ – b_pcakes Nov 7 '16 at 20:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.