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Im going to NYU as well this fall! Excited to hopefully meet you man! Keep up the great work you are doing and never give up!

Your every thing is super

You are a revelation for humanity Dr. Soborno (Doctor to be)! Srinivasan Ramanujan would have been proud of you! Proud that you are from the Indian subcontinent 🎉

Ab karwao😊

"It is obvious that the kid absolutely doesn’t know what he is talking about. He has not published any scientific papers with rigor, so how can others call him 'Professor' or 'Dr.'? In my opinion, the main point of this farce is to get money and fame."

Super Boy ❤❤❤❤❤

তুমি আমাদের অহংকার। তোমাকে দেখলে আমি অনুপ্রেরণা পাই

From Bangladesh. mashaAllah .So proud of u Suborno

Very nice dear subrno

May all your good dreams come true, Suborno✨

This genius is real motivation ❤

Very good your work im salute you

I think something's fishy with this kid. I wish I was wrong!

🎉🎉🎉🎉 proud of you specially father👏👏

bhai ki height se judge mat karo yaar woo 24 sall ka hai adolsence ki kami ka shikar

They are phenomenal 😍 Truly priceless ❤ Lovely beautiful brothers, also! ❤

Dr Suborna Isaac Professor God gifted power man talented honesty handsham gentilman in the world

kya yar bacho ka naam kharab kar rha e?😂😂

I hate science

1 minutes silence for his neighbors kids and relatives kids 🥱😂

Proud of him really God bless buy

❤❤❤❤❤

তোমার নামের বানান ভুল কেনো একেক জায়গায় একেক ভাবে লেখো কেনো,,, তোমার সঠিক বানান কোনটি (SUBORNO OR SOBORNO) কোনটি সঠিক বানান তোমার মত legend মানুষ এর বানান ভুল করা ঠিক হয়। এই বিষয়টি গুরুত্ব পূর্ণ মনে করে সঠিক বানান কোনটি announcement করে জানিয়ে দাও পৃথিবীকে। শুভ কামনা রইলো সুবর্ণ

Dr Suborna Isaac Professor God gifted talented handsome gentilman in the world super God bless

I am also proud of you

❤ from nepal

Goosebumps

Ye kon hai bhai?

Proud Of Bangladeshi All People

This is the Supreme way of learning. Great both of you and appreciate to their father for this idea 💡

Congratulations Shuborno on your great success. Keep it up, glow your parents face, your country and do the best for nice universe 🙏 from Durgapur West Bengal India ❤

❤😊🎉

You have come one day in our school SFS school Saint Francis disease

Me who don't know nothing at that age.And you are saying embracing moment no bro you are doing great

Proud 👏🎉🎉🎉🎉🎉🎉🎉🎉🎉of fill

Jhaatt ka professor

young sheldonn

😂❤

He is Bangladeshi.proud are all❤

He is a Bangladeshi boy

Only professor who completed high school graduation after being professor.

I love you My india

Love you darling❤🎉 I think I've found a hero and a role model for my star boy (son) who's present life is similar to yours early life.

\int_{0}^{\infty} e^{-x^2} , dx = \frac{\sqrt{\pi}}{2} ]

3 The best proof I've seen is a topological one, using a few theorems about different topologies. Namely: 1) In an order topology, [a,b] is a compact set. 2) The order topology on R is equivalent to the metric induced topology on R . 3) Closed subsets of compact sets are compact. Compact subsets of Hausdorff spaces are closed, and in the metric topology we can take Bn(0) , the ball of radius n around 0, and since ∪∞n=1Bn(0)=Rm , we can choose a finite subcover, and so it is a bounded set. That proves necessity. To prove sufficiency we simply note the following.Let K be a closed, bounded subset of Rn . Then K⊆[−M,M]n is actually compact, as [−M,M] is compact by 1), and since the order and metric topologies are equivalent, it is compact in the metric topology. By Tychonoff's Theorem(not even necessary, finite products of compact sets are compact), we have [−M,M]n is also compact. Closed subsets of compact sets are compact, and K is a closed subset of [−M,M] , so K is compact, proving sufficiency. One can use the following lemma to prove it: Let (Rn,d) be a Euclidean space with either the Euclidean metric or the taxicab metric. Then every bounded sequence in (Rn,d) has at least one convergent subsequence. 0 Use the following 2 propositions as step-stones. (1): Sequences in Rn converge with respect to the sup norm metric, the taxicab metric, or the Euclidean metric if and only if it's component-sequences converge individually in R . (2): (Heine-Borel Theorem for the line) X⊆R is closed and bounded if and only if any sequence in X has a convergent subsequence converging to a point in X .By contradiction, we can show that ∀1≤i≤n , the projection of E on the ith dimension Ei={xi∈R:xi is the ith component of some x∈E} is a bounded and closed subset of R . Take a sequence