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This is the proof I have obtained to show that any ideal of a Banach algebra is closed:

If $\cal B$ is a Banach algebra, then let $I$ be an ideal (bilateral). Let $\{y_n\}_{n\in{\mathbb N}}$ a sequence of elements of $I$ converging to $y$ in $\cal B$, then $y$ must be in $I$. Indeed, if it is not in $I$, then there must exists an element $x$ in the complement of $I$ such that $xy$ is not in $I$. Let $\{x_n\}_{n\in{\mathbb N}}$ a sequence in the complement of $I$ converging to $x$. Because such sequence is an arbitrary sequence, the sequence $\{x_n y \}_{n\in{\mathbb N}}$ can be chosen to be entirely contained in the complement of $I$ and its elements to have distance from $I$ greater than $0$ for $n$ finite. Moreover the sequences $\{x_n y_m\}_{m\in{\mathbb N}}$ for $n$ fixed, are entirely contained in $I$ and they converge to the elements $x_n y$ for $n\in{\mathbb N}$. This is an absurd because the elements $x_n y$ have been chosen to have distance from $I$ greater than $0$ for $n$ finite. Therefore $y$ must belong to $I$ which is therefore closed.

Where I am wrong?

stkcpc
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    I think sentence 5 is not necessarily true without knowing beforehand the result you're trying to prove. – Nick Mar 10 '24 at 21:37
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    For an example of a non-closed ideal, take the Banach algebra $C[0,1]$ of continuous real-valued functions on $[0,1]$ with pointwise operations, and the ideal $J$ consisting of functions $f \in C[0,1]$ such that $f = 0$ on some interval $[0,\epsilon]$ with $\epsilon > 0$ ($\epsilon$ depending on $f$). So where does your proof go wrong? – Robert Israel Mar 10 '24 at 21:45
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    There’s no reason you can choose $x_ny$ to have positive distance from $I$. While it is certainly possible to do so for $x_n$, $x_ny$ is a different story. – David Gao Mar 10 '24 at 21:54
  • To reply on the sentence 5 and the problem of finite distances in that sentence. Because $x_n$ is arbitrary in the complement of $I$ for a fixed $n$, it can be any point of the complement of $I$. If it does not exist an element $\xi$ in the complement of $I$ such that $\xi y $ has distance greater than zero from $I$, then I should conclude that the whole exterior of $I$ is mapped continuously on the boundary of $I$ by the product map $\mu(\cdot,y):\xi\in{\cal B}\rightarrow \mu(\xi,y)$ (because the product map is continuous for normed algebras). But, is this possible? – stkcpc Mar 10 '24 at 22:19
  • Related: https://math.stackexchange.com/questions/2523589 https://math.stackexchange.com/questions/1158397 Test your "proof" on one of these counterexamples to find your flaw. – Anne Bauval Mar 10 '24 at 22:51
  • @stkcpc Yes, that is possible. Hanno's answer contains examples where the exterior of $I$ are contained in the closure of $I$ (that is, $I$ is dense), but you could have examples where that is not the case, say $c_{00}$ (finitely supported sequences) in $l^\infty$. Then for any $y \in c_0 \setminus c_{00}$, $x \mapsto xy$ is indeed a continuous map from $l^\infty$ to $c_0$, which is the closure of $c_{00}$. – David Gao Mar 11 '24 at 07:38
  • In fact, the closure of any ideal is an ideal, so this is always the case. – David Gao Mar 11 '24 at 07:40
  • Ok, thanks! I understand better the problem now! – stkcpc Mar 11 '24 at 18:53

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Your argument does not work in the ubiquitous case when the ideal $I\triangleleft\cal B\,$ is a dense subset, since then each $b\in\cal B$ has distance zero from $I$.

Concrete examples are the ideal of finite-rank operators $\cal F(\mathcal H)\triangleleft\cal K(\mathcal H)\,$ in the compact operators on a (-n infinite-dimensional) separable Hilbert space $\mathcal H$.
Or simpler, the commutative Banach algebra $c_0 = \left\{(x_n)_{n\in\mathbb N} \mid x_n \text{ goes to zero}\right\}$ and its ideal of sequences having finite support.
(Thus, restriction from the first mentioned realm of noncommutativity to commutative subalgebras.)

Hanno
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