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

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?

• 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 $\{(z_n)_{n=0}^\infty: \sup_{n\in\mathbb{N}} |z_n|<\infty\}$ Commented Aug 1, 2020 at 20:02

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

• Thank you! I was not sure how to write down the proof but this clears it up. Commented Aug 2, 2020 at 17:44
• Glad to be of help! Commented Aug 2, 2020 at 18:09

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 $$\mathbb C^\infty$$.

• Why is $\mathbb{C}^\infty$ a problem?
– Zuy
Commented Aug 1, 2020 at 21:00
• It isn't. It's just less common (at least in Functional Analysis) because you cannot put a norm to it. Commented Aug 1, 2020 at 21:02
• 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 Commented Aug 1, 2020 at 21:05
• Usually, $$\|f\|_\infty=\sup\{|f(t):\ t\}.$$ In measure theory things are slightly more complicated, but the spirit is the same. Commented Aug 1, 2020 at 21:55