Real Analysis | Closed Sets

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[TheMauror]

17:55 that was not a good place to stop :(

[goodplacetostop]

Probably the good place to stop of this video disappeared in the black hole from the video about isolated points 😂

[parameshwarhazra]

During quarantine, Michael penn really got savage and worked harder to teach every single thing of pure mathematics. Thanks Michael. Keep it up. With love from India

[johnalley]

!!!! Intuition !!!!

Nicely - nay, masterfully - done.

You rock.

[sanjursan]

This is a very good lecture. Easy to take notes, and not miss anything. The economy of his presentation is awesome. I am self teaching a look at Modern Analysis using Simmons as lectures to the more comprehensive, but drier Bachmann and Narici, so these lectures by Prof Penn are priceless to me.
Remember folks, do the math!

[anitapandey]

Analysis is too much rigorous but interesting. Thanks for these lectures Mr. Penn.

[axelperezmachado]

Just for fun and completeness, the last proof could have finished as:

let x in V_e(a)
if x=a, because a is in A then x is in A
if x =/= a
then x is in V_e(a) \ {a}
then x is not in A^c (if it where, then (V_e(a) \ {a}) intersect A^c would not be empty)
then x is in A
and thus V_e(a) is completely inside A

[henrywoo]

Great video delivered by a great professor . Thank you Professor !

[elgourmetdotcom]

Oh, what happened? Does this compensate the extra minutes from the previous video? 😂 love this RA course

[wtt]

Thanks so much for this great video .

[phukinho]

7:52 I can prove that b is in the epsilon neighbourhood of x if epsilon > x-b but not if epsilon >= x-b :(

[billh]

You want to prove that [b, c] is closed. According to your definition, you don't need to prove that x in [b, c] implies x is a limit point. Your definition only requires: x is a limit point of [b, c] implies x in [b, c].

[Zealot]

I may missing some fundimental part here. Suppose A = { 0 and 1/n | n is positive integer }, from the definition, A is closed, because it contains all limit points, which is only one point "0". But I'm pretty sure A completement is open.

[urieldaboamorte]

being inactive during quarentine? covid's got nothing on my boy

[abhiuser-zjmmyp]

Please do videos about elliptic integrals

[ichkaodko]

is limit point and boundary point same thing or different?

[RandomBurfness]

How is this real analysis if there are no metric spaces? :P

[spencerpencer]

pardon, but it appears to me that your proof of the first claim contains some redundancies. if you prove that x is not in [b, c] implies it is not a limit point (which you do) it necessarily follows that the only limit points of [b, c] are contained in [b, c] and you are done. you do not need to show that every point in [b, c] is a limit point even though it is true

[CousinoMacul]

I don't think that your proof takes into account A being the set of real numbers and A compliment being the empty set.

[jamescollis]

(0, 1] is open as it doesn't contain 0, surely the complement is also open as it doesn't contain 1? What am I missing?