@@Sorya-gf7qw If x is rational, the union of A and B are not the whole set of rational numbers Q, because the rational number x is missing. So for the argument in 7:15, you need that x is irrational.
great video! Do you have a video on locally connectedness? There seem to be these interesting sets that are connected but not locally path connected? I am not familiar with the english terminology, but basicly there is this set, where a component can't be seperated as an open subset, but you can't have a path leading to it at the same time. It's kind of weird but also super fascinating.
A connected set is a set that cannot be partitioned into two nonempty subsets which are open in the relative topology induced on the set. But seems that [1,2]U[3,4] is disconnected but also cannot be the union of two nonempty open sets.
Real analysis is the study of spaces of real numbers \R^n, including properties of subsets of \R^n and real-valued functions on them. The most elementary form of real analysis is real-valued calculus, which students often study in high school. At the college level, it includes vector calculus and express courses on real analysis, and it dovetails into fields like measure theory and functional analysis.
Additionally, it is also the study of the properties of different sets such as compactness, closure, completeness and many more. This then extends into a topic known as metric spaces which in turn generalises to topological spaces.
Sure it does. if a,b are both in {5}, then both a=5 and b=5, so the interval (a,b) is the empty set, and the supposition that (if c in (a,b), then c in E) is vacuously true, as (a,b) is empty.
8:20
The closure of A and B should both include x. But doesn’t change the argument.
I was wondering exactly that, thanks for clarifying!
Yes, because of course x will be a limit point of both A and B
Sir in that case I wonder why take x an irrational number, why won't a rational x work ?
@@Sorya-gf7qw If x is rational, the union of A and B are not the whole set of rational numbers Q, because the rational number x is missing. So for the argument in 7:15, you need that x is irrational.
Thank you jacked math man.
Really love that video series of analysis!
The closure of A in the claim of set Q is disconnected should be (-∞,x], right?
Do you have a playlist of all of your real analysis talks? Would love to watch them in order.
czcams.com/play/PL22w63XsKjqxqaF-Q7MSyeSG1W1_xaQoS.html
great video! Do you have a video on locally connectedness? There seem to be these interesting sets that are connected but not locally path connected? I am not familiar with the english terminology, but basicly there is this set, where a component can't be seperated as an open subset, but you can't have a path leading to it at the same time. It's kind of weird but also super fascinating.
Wish You were my professor in my undergrad
18:58
a typical introductory course of calculus is actually a course on the topology of metric spaces in disguise
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change my mind
3:20 ERROR
A connected set is a set that cannot be partitioned into two nonempty subsets which are open in the relative topology induced on the set. But seems that [1,2]U[3,4] is disconnected but also cannot be the union of two nonempty open sets.
I see... It is something about the open set of the relative subspace.
@VeryEvilPettingZoo Thank you so much!!!
Can someone please please please tell me what is real analysis. What is it connected with. And when do students study about it.
Real analysis is the study of spaces of real numbers \R^n, including properties of subsets of \R^n and real-valued functions on them. The most elementary form of real analysis is real-valued calculus, which students often study in high school. At the college level, it includes vector calculus and express courses on real analysis, and it dovetails into fields like measure theory and functional analysis.
Additionally, it is also the study of the properties of different sets such as compactness, closure, completeness and many more. This then extends into a topic known as metric spaces which in turn generalises to topological spaces.
A single element like {5} would be connected. The proof would not consider this case.
Sure it does. if a,b are both in {5}, then both a=5 and b=5, so the interval (a,b) is the empty set, and the supposition that (if c in (a,b), then c in E) is vacuously true, as (a,b) is empty.
Sir u know hindi..?
Why you even care about that
If you think he’s going to start making hindi videos I think you need to reconsider brain in general. He’s American with an English speaking audience.
am I the first hehe