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I have been reading about Fredholm operators and came across these notes.

My question is about Lemma 16.17 on the second page. This lemma states the following:

Lemma 16.17: Let $X$ and $Y$ be Banach spaces and $T:X \rightarrow Y$ a bounded linear map. The following are equivalent:

  1. $\text{Ker}(T)$ is finite dimensional and $\text{Im}(T)$ is closed
  2. Every bounded sequence $\left \{ x_{i} \right \} \subset X$ with $Tx_{i}$ convergent has a convergent subsequence.

I am having a hard time seeing the reverse direction ($2. \implies 1.$). In particular I am interested in 2. implying $\text{Ker}(T)$ is finite dimensional (the image being closed was covered in another question). It seems to me like this is connected to the Bolzano-Weierstrass property which only holds in finite dimensional spaces, but I am not sure how or where the convergence of $Tx_{i}$ comes in.

Any help would be appreciated.

user273331
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    Take a bounded sequence in $\ker T$ and use 2. to show that is has a convergent subsequence. Conclude that the (relative) closed unit ball of $\ker T$ is compact and thus the kernel is of finite dimension. – Evangelopoulos Foivos Jun 07 '21 at 05:14
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    Ah thank you. All $Tx_{i} = 0$ so all subsequences converge to the same point. So the $Tx_{i}$ condition in 2. is automatically satisfied. Appreciate the help. – user273331 Jun 07 '21 at 06:15

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