With the standard topology on R,which one of the sets is open in R?

- anonymous

With the standard topology on R,which one of the sets is open in R?

- schrodinger

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- anonymous

\[{x: \frac{1}{2}\ < |x| < 1} \]

- anonymous

i fink this is the answer but i have other options

- zzr0ck3r

so why is this open? Can you write it in interval notation?

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## More answers

- anonymous

(1/2,1)

- zzr0ck3r

but the absolute value sign means you also get (-1,-1/2). Do you see that?
1/2 < |-0.8|<1

- anonymous

ok,

- zzr0ck3r

So the set, in interval notation, is \[(-1,- \dfrac{1}{2})\cup (\dfrac{1}{2}, 1)\]

- anonymous

please why -0.8?

- zzr0ck3r

I was just showing you that you get negative number in there as well

- zzr0ck3r

\(\{x\mid \frac{1}{2}< |x|<1\}=(-1,-\frac{1}{2})\cup (\frac{1}{2}, 1)\)

- anonymous

ok. now i got it. because of the absolute value .

- anonymous

should i post the other options?

- anonymous

\[{x: \frac{1}{2} < |x|\leqslant1} \]

- anonymous

now this is neither open or close in R . right ?

- zzr0ck3r

Can you write it in interval notation?

- anonymous

please don't shout on me. i will try

- zzr0ck3r

what?

- anonymous

i mean please do not get angry if i fail

- zzr0ck3r

I don't even know how to respond to that.

- anonymous

what if i say it can not be written as union of open sets

- anonymous

(-1,-1/2)u(1/2,1]

- anonymous

you there?

- zzr0ck3r

almost, the -1 gets a [ also

- anonymous

i thought as much

- zzr0ck3r

\([-1, -0.4)\cup(0.5,1]\)

- zzr0ck3r

|-1|<=1

- zzr0ck3r

ok what is a closed set?

- zzr0ck3r

you gone?

- 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

no. i am here,network is getting bad

- anonymous

- 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

Ok, we can go with boundary, does this set contain all of its boundary points? If not, which 2 does it not contain?

- 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

am sorry sir. my network is really poor

- 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

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

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

- zzr0ck3r

yes, unless I did not see all the options

- anonymous

ok sir

- anonymous

let me close this question and ask another . can i?

- zzr0ck3r

of course

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