Let
X=(a,b,c,d,e)andτ=(X,ϕ,[a],[c,d],[a,c,d],[b,c,d,e]).LetA=[a,c]
, then set
A′of limit points of A
is given by

- anonymous

- chestercat

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- anonymous

A′=(b,c,e)
A′=(b,d,e)
A′=(b,e)
A′=X

- anonymous

- anonymous

i think option B

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## More answers

- anonymous

- anonymous

am i correct ?

- zzr0ck3r

Do you mean interval [ ] ?

- zzr0ck3r

or { }

- zzr0ck3r

do you mean the empty set \(\emptyset\) and not that weird thing?

- anonymous

i think {} and that weird thing is empty set

- zzr0ck3r

ok can you edit it to make sense?

- zzr0ck3r

these things mean different things...

- anonymous

X=(a,b,c,d,e)andτ=(X,ϕ,{a},{c,d},{a,c,d},{b,c,d,e}).LetA={a,c}
, then set
A′ of limit points of A
is given by

- anonymous

- zzr0ck3r

why do you think B?

- anonymous

because that is A complement with X

- zzr0ck3r

So you think that the set of limit points is defined to be the compliment?

- anonymous

please i will be happy if you explain the concept with you

- zzr0ck3r

I already have.

- zzr0ck3r

What is a limit point?

- anonymous

it is the boundary points or closed interval points

- anonymous

- zzr0ck3r

So you did not read a thing I wrote on the last problem?

- zzr0ck3r

Go back and read that and then tell me what a limit point is

- anonymous

a point for which every neighborhood contains at least one point belonging to a given set.

- zzr0ck3r

Also, when you reply to something I have commented on, I get tagged and see it. There is no reason to tag me more than once on a post.

- zzr0ck3r

ok, so what are the open sets containing b?

- anonymous

,{b,c,d,e}

- zzr0ck3r

does that intersect A? in other words, does that share any points with A?

- anonymous

yes c

- zzr0ck3r

ok, what are the open sets containing d?

- anonymous

{c,d},{a,c,d},{b,c,d,e}

- zzr0ck3r

I shuold be saying intersect A at some other point...

- zzr0ck3r

ok do all of those sets intersect A at some other point other than d?

- zzr0ck3r

why are you using caps?

- zzr0ck3r

d is not in A... try again

- anonymous

sorry . 'a' for the first and 'c' for the second and third

- zzr0ck3r

right

- zzr0ck3r

ok and with e we get the same thing as with b right?

- anonymous

yes

- zzr0ck3r

So your option is looking good, but maybe some other point is also a limit point.
What open sets contain a?

- zzr0ck3r

Better yet, do all of the open sets that contain a also intersect A at some point other than a?

- anonymous

{a},{a,c,d}

- zzr0ck3r

do all of the open sets that contain a also intersect A at some point other than a?

- anonymous

yes

- anonymous

but not {a}

- zzr0ck3r

right so you mean no then?

- anonymous

yes

- zzr0ck3r

then a is not a limit point
what about c?

- zzr0ck3r

is c a limit point?

- anonymous

yes ,{a,c,d} because it intersects A

- zzr0ck3r

is that the only open set that contains c?
Remember, we only have to find one that does not intersect at some point other than c

- anonymous

,{c,d},{a,c,d},{b,c,d,e} contain c

- zzr0ck3r

do all of those intersect A at some point different than c?

- anonymous

only {a,c,d}

- zzr0ck3r

so there is at least one that does not?

- zzr0ck3r

then c is NOT a limit point.
So yes your option B is the answer, but not for the reason you listed :)

- anonymous

ok. thanks . i have to view this page more often to understand this

- anonymous

Let
A=(0,1]⋃2
be a subset of
R

- anonymous

what are the limit points

- zzr0ck3r

remember to show something is not a limit points you need to show that it has an open set around it that intersects the set at some point OTHER than the point in question. I did not stress that in the last post.

- anonymous

ok thanks

- zzr0ck3r

For a subset \(A\subset X\) a limit point \(x\) of \(A\) is a point in \(X\) (not necessarily in \(A\)) such that all open sets containing it intersect \(A\) at some point other than \(x\).

- anonymous

i think 0 and 1

- anonymous

ok

- zzr0ck3r

you think what 0 and 1?

- anonymous

Let
A=(0,1]⋃2
be a subset of
R

- zzr0ck3r

done

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