What do you think about bartle - elements of real analysis?

Yes, no problem: look up "discrete topology". Another useful one for your topology toolkit is the "trivial topology". Both are good for testing proposed theorems and finding counter examples. There is a definition of what makes something a metric though and that definition garantees that you get a proper topology out of it. But not every definition of "distance" is a metric.

Oh no 🌚

I had the same question about Interior Point Example 2. If we limit ourselves to considering only rationals then any ball B around x∈Q° would contain all the elements of Q within the bounds of B. Are interior points strictly defined to be Reals?

@@Sahanie Generally surely no, but I believe the topology over the reals induced by the Euclidean metric was implicitly assumed in the second example at the end. Considering the discrete topology (power set) over just the rational numbers would make each rational number an interior point.

Yes. this is true for every definition . 'If' is commonly used for 'Iff' in definitions

A set can be both open and closed, so the empty set is open but it’s also closed because it’s complement is S which is open

Not true

@@drpeyam Thank you

Even technically you are correct. The "boundary" is defined as the closure (smallest containing closed set) minus the interior (there are other equivalent ways to define it as well). See e.g. en.wikipedia.org/wiki/Boundary_(topology). But a set is open if and only if it is equal to its interior. So if a set does not contain any of its boundary points, it is no larger than its interior and that means that it is equal to its interior and therefore open. And if it is open then it is equal to its interior, which is excluded from its boundary, so it does not contain any of its boundary points. I guess that is exactly what you are saying. E.g. the closure of Q (as a subset of R) is R, so the boundary is R (and Q is not open in R). The closure of Q as a subset of Q is Q, so that the boundary is empty (Q\Q) (and Q is open in Q). Happy to be proven wrong and see a counter example. Note that while this introduction video to open sets is understandably based on a metric (or norm if you like), that still defines a topology and the more abstract topological considerations and facts apply equally well to it.

@@alexfekken7599 Yes that is what I meant. Thank you very much for your explanation. My comment was pretty vague 😅

@@alexfekken7599 If we consider the normal open-ball-topology induced by the Euclidean metric over the reals, then for the semi-open interval E = [0,1), its boundary {0,1} is not included in E but the set E is also not an open set with respect to the space's topology. EDIT: I think the answer to this depends on whether we mean "E does not include its (entire) boundary" or "E does not include any points that belong to its boundary"

It’s open

Check out my playlist

@@drpeyam chi Migi baba there’s not topology playlist

Metric spaces playlist

@@drpeyam I mean like on connectedness compactnesss path connected topological invariants..

Literally that playlist

Correct: you need to specify the enclosing set and its topology (just like you need to specify domains and codomains when you talk about functions). Q as a subset of itself with the usual topology is open (and closed). As a subset of R with the usual toplogy it is not open (nor is it closed). (0,1) as a subset of R with the usual toplogy is open. As a subset of C with the usual toplogy it is not open (nor is it closed).

The metric space. Also what is the 33 in your name?

@@drpeyam It's my roll number in school 😆

That's true but he hadn't defined closed sets yet so he's leaving it for another video

Ah I wish it was said before, it would have put me out of my misery...I saw the contradiction between what was said at 9:30 and then 11:50 when Dr Peyam derives the converging zero set to be closed. Your comment makes it clear. Thanks!

We’re saying there is an r such that this exists, not that for every r this is true. And your 8.999 argument assumes that 9 is in your set which it isn’t

@@drpeyam thank you for your reply, but could you please explain to me why 9 wouldn’t be part of my set?

​​​@@amadouz6475 because no matter how much small 'r' u take like 0.00000001, 9-r is still outside this interval i think dr peyam got ur point its not like definition of limit we don't need every r>0 we just need to check if there exists such r I am also learning if I am wrong then please correct me

He means S as in a general set S. This is so that you can define topology for all kinda sets. But yes you could take S to be ℝ

@@helloitsme7553 why is S, a general set open?

@@Happy_Abe You define it that way. You act like S is the biggest set lets say, and you wanna know which subsets of S are open. This then coincides with the idea of openness where balls are always contained, if S is the biggest set, any ball is automatically in S so S is open.

@@Happy_Abe This S is the set that's part of the metric space you're working in. That is, if you're working in a metric space (S, d), that S is open in that metric space. On the other hand, any particular set is not open in some other metric space. For example, R is not open in the metric space (C, d(x,y)=|x-y|) (that is, the metric space you'd expect for the complex numbers), because 0 is in R, and, for any r>0, |(0+(r/2)i) - 0| < r but 0+(r/2)i is not in R.

It’s an acquired taste, but check out the playlist