At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

i keep getting int and bd wrong

I wish I could help, dunno enough about topology yet

ok thank you though :)

I can help reason it out if you tell me definitions

Seems like interior is just stripping away the boundery.

then x is called a boundary point of S.

You understand what neighborhood is?

and yes i understand what neighborhood is

i would have guessed the ans for b) to be 3 or something

Despite not understanding neighborhood rigorously, that is my reasoning for b)

I mean for a)

And it applies to c) and d) apparently

@manjuthottam make sense?

so the answer is focusing on rational numbers?

Okay, @manjuthottam do you know what continuity is?

is it just that the limit exists?

yeah the limit exists at the function

But actually, let's talk about neighborhoods a bit ok?

ok so neighborhood is x in R. while there is an epsilon greater than zero where |x-y|< epsilon?

Yeah, notice how that looks kinda like a limit....
Anyway...

Consider 1/2, 1/3, and 1/4
Do they have a neighborhood?

i'm guessing no?

at least its not that specific of a number

So can you understand how these three points would be boundary points?

In this way, neighborhoods are analogous to continuity.

hmm ok

Does my analogy sound consistent to you understanding of what a neighborhood is?

yes it does

yes i see

thus it is not in our set

hmm ok that makes sense!

This is sort of a proof by contradiction as to why there can't be any neighborhoods.

ok, so then based on the defn of int and bd, we have to first see about continuity and neighborhood

if it is continuous then we can see that int and bd are not an empty set

The continuity thing was more to help me understand neighborhoods than it was to help you. =)

=D kk but it helped me learn too! lol!

I think what you need to understand is that rational numbers do NOT make any neighborhoods.

you'll always be able to find some irrational number between two rational numbers.

ok

ok so even union of sets

Yeah, it should work for unions of sets well too
Just not for intersections of sets.

so like e)

automatically the int is empty set

So for intersections... just compute the intersection. It shouldn't be too bad.

yeah

You'd have to give me some \(\epsilon>0\) but clearly \(2+\epsilon\) is NOT in the set \(\{2\}\)

kk thank you so much for helping me !!!!