The closed graph theorem on wikipedia is given as follows:
I'm not understanding how this isn't a tautology. Isn't the definition of continuity that $\{x_n\} \rightarrow x \implies \{f(x_n)\} \rightarrow f(x)$?
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The definition of continuity is: If $U\subseteq Y$ is open, then $T^{-1}(U)$ is open in $X$ – Hagen von Eitzen Jun 19 '21 at 21:15
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sure but another definition is that assuming that a sequence converges in the domain, the image of that sequence converges in the range (with the appropriate assumptions on the spaces we're considering). – yoshi Jun 19 '21 at 21:19
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1@yoshi: Not all forms of convergence are characterized by sequences. – Jacob Manaker Jun 20 '21 at 07:39
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There are many, many equivalent definitions for continuity. What you mention is one of them, at least when we are talking for first countable topological spaces. The interesting part in the CGT are numbers 4 and 5. Basically the CGT tells you this: start with a linear operator $T:X\to Y$ between Banach spaces and say you want to check whether this is continuous. With the definition you provided, you would have to take a sequence $\{x_n\}\subset X$ that converges to some $x\in X$ (i.e. $x_n\to x$) and under ONLY this assumption, you have to show that $Tx_n\to Tx$.
Basically, number 4 of the CGT tells you "hey, it's okay, you can ALSO assume that $Tx_n$ converges to some $y\in Y$, i.e. you can ALSO assume that $\{Tx_n\}$ is convergent. You only need to show that $y=Tx$."
You may not see why this is a tremendous help, but when applying it you will understand its magnitude. Here are some interesting applications:
Here is a suggestion: try to prove (1) without using CGT. Can you do it? (well, even if you can, I bet the answer will be 10 times harder than a simple application of the CGT).
Just dropped in
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I want to add, that the number 2 is also important. It leads us straight to the open mapping and the inverse mapping theorems. – Timur Bakiev Jun 19 '21 at 21:43
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@TimurBakiev Yes, of course. Feel free to edit my post if you want:) – Just dropped in Jun 19 '21 at 21:44
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1$(1)$ is not too bad if you allow Uniform boundedness principle, and you don't need that the spaces are Banach. Assume $T$ is unbounded so there is a sequence $(x_n)_n$ such that $|x_n|=1$ and $|Tx_n| \ge n^2$ for all $n \in \Bbb{N}$. Define $y_n := \frac{x_n}n$ so that $y_n \to 0$ but $|Ty_n| \ge n, \forall n \in \Bbb{N}$. Then by the continuity assumption we have $f(Ty_n) \to 0$ for all $f \in Y'$ which means $Ty_n \to 0$ weakly. But weakly convergent sequences are bounded which contradicts $|Ty_n| \ge n, \forall n \in \Bbb{N}$. Therefore $T$ is bounded. – mechanodroid Jun 19 '21 at 22:18
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@mechanodroid yeah, I was afraid that someone would see a u.b.p alternative. Nice one! Well, u.b.p is a result of the same strength as c.g.t so I guess the point I was trying to make remains:) – Just dropped in Jun 19 '21 at 22:22