I am trying to prove that every vector space $X$ has a norm. I have some silly questions, but it's better to ask them now instead of later. I think I'm having a bit of trouble getting intuition about basis in infinite dimensional spaces.

Fix a Hamel basis ${\cal B} = ({\bf e}_i)_{i \in I}$. Then for all ${\bf x} \in X$, we write: $${\bf x} = \sum_{i \in F}a_i{\bf e}_i,$$ for some $F \subset I$ finite. I understand that this combination is unique in the sense that: $$\sum_{i \in F_1}a_i{\bf e}_i = \sum_{i \in F_2}b_i{\bf e}_i,$$ for some $F_1,F_2 \subset I$ finite, implies that $a_i = b_i$ for all $i \in F_1 \cap F_2$ and $a_i = 0$ for all $i \in F_1 \setminus F_2$, and $b_i = 0$ for all $i \in F_2 \setminus F_1$.

Does this mean that $F_1 = F_2$?

Assuming that yes, although I'm not sure, the idea would be to define some kind of max norm, which would be well-defined: $$\|{\bf x}\| = \max\{|a_i| \mid i \in F \}.$$ This idea seems good, it even showed up in another answer. The properties $\|{\bf x}\| \geq 0$ for all ${\bf x}\in X$, $\|{\bf x}\| = 0 \implies {\bf x}=0$ and $\|\lambda{\bf x}\| = |\lambda|\|{\bf x}\|$ are all clear. I'm having trouble getting the triangle inequality. How to make a sum here is a bit confuse to me. If ${\bf x},{\bf y}\in X$, then there are $F_1,F_2 \subset I$ finite such that: $${\bf x} = \sum_{i \in F_1}a_i{\bf e}_i, \quad \text{and}\quad {\bf y}=\sum_{i\in F_2}b_i{\bf e}_i,$$ so that $${\bf x}+{\bf y} = \sum_{i \in F_1 \setminus F_2}x_i{\bf e}_i + \sum_{i\in F_1 \cap F_2}(a_i+b_i){\bf e}_i + \sum_{i \in F_2 \setminus F_1}b_i{\bf e}_i.$$ I wanted to write this as $\sum_{i \in \text{ something}}c_i{\bf e}_i,$ but the only thing I could think of was: $$\sum_{i \in F_1 \cup F_2}c_i{\bf e}_i, \quad c_i = \begin{cases} a_i, \text{ if }i \in F_1 \setminus F_2 \\ a_i+b_i, \text{ if }i \in F_1 \cap F_2 \\ b_i, \text{ if }i \in F_2 \setminus F_1\end{cases}$$

But again:

Is this combination unique in the sense that the only possible combination for the vector ${\bf x}+{\bf y}$ will be indexed by $F_1\cup F_2$?

I think I am overcomplicating things. I can get the triangle inequality with this, it seems, but things don't look well-defined enough for me. Can someone address these questions and give me a small explanation about it? Thanks.

Edit: to confirm what I understood from gerw's answer: since the combination is unique, if I write $${\bf x}=∑_{i∈F_1}a_i{\bf e}_i=∑_{i∈F_2}b_i{\bf e}_i,$$ for $F_1,F_2⊂I$, finite sets, then: $\max\{|a_i|∣i∈F_1\}=\max\{|b_i|∣i∈F_2\}$, and this ensures that $\|\cdot\|$ is well defined, right?

  1. No, you do not necessarily have $F_1 = F_2$. But as you have already shown the coefficiencts of the indices in $(F_1 \setminus F_2) \cup (F_2 \setminus F_1)$ are zero.

  2. This combination is unique, up to zero coefficients, cf. 1.

  • $\begingroup$ Thanks for the answer. Can you please confirm if I understood everything ok? Since the combination is unique, if I write $${\bf x} = \sum_{i \in F_1}a_i{\bf e}_i = \sum_{i \in F_2}b_i{\bf e}_i,$$ for $F_1,F_2 \subset I$, finite sets, then: $$\max\{ |a_i| \mid i \in F_1 \} = \max\{ |b_i| \mid i \in F_2 \},$$ and this ensures that $\|\cdot\|$ is well defined, right? $\endgroup$ – Ivo Terek May 29 '15 at 19:00
  • $\begingroup$ Yes, you are right. $\endgroup$ – gerw May 29 '15 at 19:24
  • $\begingroup$ Ok. Thanks again! $\endgroup$ – Ivo Terek May 29 '15 at 19:25

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