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?
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.
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 $U$.
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.
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 $\mathbb{C}^{\infty}$.
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 $\mathbb C^\infty$.
