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In Axler's Linear Algebra Done Right, they set an example for a subspace:

The set of all sequences of complex numbers with limit 0 is a subspace of $\mathbb{C}^{\infty}$,

where $\mathbb{C}^{\infty}$ denotes the vector space of complex sequences over $\mathbb{C}$.

How can I interpret the ¨with limit 0¨ part? does it mean, looking at an element of the subspace as a function f(z), that

$$\lim_{z\rightarrow z_o} f(z) = 0$$ ?

How can we go on about to prove sub set of functions is a subspace?

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  • $\begingroup$ You're probably looking at the set $\{(z_n)_{n=0}^\infty: \lim_{n\to\infty} z_n=0\}$. You want to show that this is a subspace of $\{(z_n)_{n=0}^\infty: \sup_{n\in\mathbb{N}} |z_n|<\infty\}$ $\endgroup$
    – Reveillark
    Commented Aug 1, 2020 at 20:02

2 Answers 2

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Let's put a name to our reputed subspace, say $U$. Then $U$ is the set of all complex sequences $(z_{n})_{n=0}^{\infty}$ such that $$ \lim_{n \rightarrow \infty} z_{n} = 0. $$ Let us now prove that $U$ is a subspace.

  1. Remember that the additive identity of our vector space $\mathbb{C}^{\infty}$ is the sequence whose terms are all zero: $(0, 0, 0, 0, \dots)$. It is indeed the case that the limit of this sequence is $0$, so it belongs to $U$.

  2. Now take two sequences $(z_{n})_{n=0}^{\infty}$ and $(w_{n})_{n=0}^{\infty}$ in $U$. Then $$ \lim_{n \rightarrow \infty} (z_{n} + w_{n}) = \lim_{n \rightarrow \infty} z_{n} + \lim_{n \rightarrow \infty} w_{n} = 0 + 0 = 0, $$ so the sequence $(z_{n} + w_{n})_{n=0}^{\infty}$ is in $U$. This shows that $U$ is closed under addition.

  3. Now let $\lambda$ be an arbitrary complex number. We see that $$ \lim_{n \rightarrow \infty} \lambda z_{n} = \lambda \lim_{n \rightarrow \infty} z_{n} = \lambda \cdot 0 = 0, $$ so the sequence $(\lambda z_{n})_{n=0}^{\infty}$ is in $U$. This shows that $U$ is closed under scalar multiplication.

We can therefore conclude that $U$ is a subspace of $\mathbb{C}^{\infty}$.

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  • $\begingroup$ Thank you! I was not sure how to write down the proof but this clears it up. $\endgroup$ Commented Aug 2, 2020 at 17:44
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    $\begingroup$ Glad to be of help! $\endgroup$ Commented Aug 2, 2020 at 18:09
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Most likely it says $\ell^\infty$, which is the set of bounded sequences. So you want to show that $$ c_0^{\vphantom0}=\{f:\mathbb N\to\mathbb C:\ \lim_{n\to\infty}f(n)=0\} $$ is a subspace of $$ \ell^\infty=\{f:\mathbb N\to\mathbb C:\ \|f\|_\infty<\infty\}. $$ Same effort to show that $c_0$ is a subspace of $\mathbb C^\infty$.

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  • $\begingroup$ Why is $\mathbb{C}^\infty$ a problem? $\endgroup$
    – Zuy
    Commented Aug 1, 2020 at 21:00
  • $\begingroup$ It isn't. It's just less common (at least in Functional Analysis) because you cannot put a norm to it. $\endgroup$ Commented Aug 1, 2020 at 21:02
  • $\begingroup$ Im sorry what does $||f||_{\infty}$ means? Im just starting on this book and this notation is a bit out of what ive ever seen $\endgroup$ Commented Aug 1, 2020 at 21:05
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    $\begingroup$ Usually, $$\|f\|_\infty=\sup\{|f(t):\ t\}.$$ In measure theory things are slightly more complicated, but the spirit is the same. $\endgroup$ Commented Aug 1, 2020 at 21:55

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