I am reading this paper by S.H Karin titled [Norm attaining operators and pseudospectrum](https://arxiv.org/pdf/1209.1218.pdf).
In page 2 he gives the definition of $l\_p$ direct sum of a family of Banach spaces as follows:
If $1\leq p< \infty$ and $\{X\_\alpha\}\_{\alpha\in\Lambda}$ is a faily of Banach spaces, then their $l\_p$ direct sum is the space
$$X=\bigg\{ x\in\prod\limits\_{\alpha\in\Lambda}X\_\alpha:\sum\limits\_{\alpha\in\Lambda}\|x\_\alpha\|^p<\infty\bigg\}$$ endowed with the norm 
 $$\|x\|=\left(\sum\limits\_{\alpha\in\Lambda}\|x\_\alpha\|^p\right)^{\frac{1}{p}}.$$
I want to ask if $X\_\alpha$ can be non zero for all $\alpha\in\Lambda$? Can $\Lambda$ be an uncountable set?



Regarding $\ell_p$ direct sums
Yes, $X\_\alpha$ can be nonzero for uncountably many $\alpha$. Of course, each individual vector $x \in X$ is zero except for countably many coodinates, in order for $\sum\_{\alpha\in\Lambda}\|x\_\alpha\|^p<\infty$.


**added**  

Suppose (for purposes of contradiction) $x \in X$ is such that there are uncontably many $\alpha$ with $\|x\_\alpha\| \ne 0$. Then for some $\varepsilon > 0$ there are infinitely many $\alpha$ with $\|x\_\alpha\| > \varepsilon$. [Reason: in fact, $\varepsilon$ may be chosen rational, and use the fact that a countable union of finite sets is countable.]  

Let $\alpha\_1, \alpha\_2,\dots$ be such that $\|x\_{\alpha\_k}\| > \varepsilon$. Then
$$
\sum\_{\alpha\in \Lambda}\|x\_\alpha\|^p \ge \sum\_{k=1}^\infty \|x\_{\alpha\_k}\|^p > \sum\_{k=1}^\infty \varepsilon = +\infty .
$$
Thus $x \notin X$.


