anonymous
  • anonymous
With the standard topology on R,which one of the sets is open in R?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
\[{x: \frac{1}{2}\ < |x| < 1} \]
anonymous
  • anonymous
i fink this is the answer but i have other options
zzr0ck3r
  • zzr0ck3r
so why is this open? Can you write it in interval notation?

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anonymous
  • anonymous
(1/2,1)
zzr0ck3r
  • zzr0ck3r
but the absolute value sign means you also get (-1,-1/2). Do you see that? 1/2 < |-0.8|<1
anonymous
  • anonymous
ok,
zzr0ck3r
  • zzr0ck3r
So the set, in interval notation, is \[(-1,- \dfrac{1}{2})\cup (\dfrac{1}{2}, 1)\]
anonymous
  • anonymous
please why -0.8?
zzr0ck3r
  • zzr0ck3r
I was just showing you that you get negative number in there as well
zzr0ck3r
  • zzr0ck3r
\(\{x\mid \frac{1}{2}< |x|<1\}=(-1,-\frac{1}{2})\cup (\frac{1}{2}, 1)\)
anonymous
  • anonymous
ok. now i got it. because of the absolute value .
anonymous
  • anonymous
should i post the other options?
anonymous
  • anonymous
\[{x: \frac{1}{2} < |x|\leqslant1} \]
anonymous
  • anonymous
now this is neither open or close in R . right ?
zzr0ck3r
  • zzr0ck3r
Can you write it in interval notation?
anonymous
  • anonymous
please don't shout on me. i will try
zzr0ck3r
  • zzr0ck3r
what?
anonymous
  • anonymous
i mean please do not get angry if i fail
zzr0ck3r
  • zzr0ck3r
I don't even know how to respond to that.
anonymous
  • anonymous
what if i say it can not be written as union of open sets
anonymous
  • anonymous
(-1,-1/2)u(1/2,1]
anonymous
  • anonymous
you there?
zzr0ck3r
  • zzr0ck3r
almost, the -1 gets a [ also
anonymous
  • anonymous
i thought as much
zzr0ck3r
  • zzr0ck3r
\([-1, -0.4)\cup(0.5,1]\)
zzr0ck3r
  • zzr0ck3r
|-1|<=1
zzr0ck3r
  • zzr0ck3r
ok what is a closed set?
zzr0ck3r
  • zzr0ck3r
you gone?
anonymous
  • anonymous
a close set is a close interval. meaning that every set that has it boundary involve is close . because if we take a small radius around any boundary, it will not be in the set
anonymous
  • anonymous
no. i am here,network is getting bad
anonymous
  • anonymous
zzr0ck3r
  • zzr0ck3r
Ok, closed intervals are closed sets, but that is not all of them. R is both open and closed, and so is the empty set, and neither of those are closed intervals. A set is closed if its compliment is open. Do you know what a compliment is?
zzr0ck3r
  • zzr0ck3r
Ok, we can go with boundary, does this set contain all of its boundary points? If not, which 2 does it not contain?
zzr0ck3r
  • zzr0ck3r
Aight man, I am sorry about the internet, but this is taking way to long. It took about 35 minutes to get replies. I will be back tomorrow.
anonymous
  • anonymous
am sorry sir. my network is really poor
anonymous
  • anonymous
it does not contain all its boundary point. it contains only one which is 1 but does not contain 1/2. so it is half open and half close
zzr0ck3r
  • zzr0ck3r
Its boundary points are -1, -1/2, 1/2, and 1. While it is not a great idea to think of every open/closed set as an interval, loosely said, the boundary points are the endpoints of any broken up union of sets.
anonymous
  • anonymous
so, x:1/2<|x|⩽1 is neither open nor close . so is it right to say that the only open sets in R in the options given above are \[x:1/2<|x|<1 \] and the close set in the option above in R is \[x:1/2\le|x|⩽1 \] right?
anonymous
  • anonymous
zzr0ck3r
  • zzr0ck3r
yes, unless I did not see all the options
anonymous
  • anonymous
ok sir
anonymous
  • anonymous
let me close this question and ask another . can i?
zzr0ck3r
  • zzr0ck3r
of course

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