Let $X$ be a topological space. We define the weight of $X$ as
$$ \omega(X):=\min\{|\mathcal{B}|: \mathcal{B}\text{ is a base for}\ X\}. $$
Now, consider $\mathbb{I}:=[0,1]$ and $I$ an arbitrary infinite set. I need to show that $\omega(X)=|I|$ where $X:=\mathbb{I}^I$. It is easy to see that $\omega(X)\leq|I|$ because the canonical base of $X$ has size $|I|$. I have troubles with the other inequality. My idea is to find a familiy of $|I|$ open sets of $X$ such that each one of them are disjoint by pairs, but I can't get nothing. Can someone give me a hint?
Suppose for contradiction that $\mathcal{B}$ is a base for $X$ with $|\mathcal{B}|<|I|$.
Let $x\in X$. Then $\mathcal{B}_x = \{U\in \mathcal{B}\mid x\in U\}$ is a neighborhood base for $x$, and $|\mathcal{B}_x|<|I|$.
For any $U\in \mathcal{B}_x$, there is some standard basic open set $V$ such that $x\in V\subseteq U$. Then there is a finite set $J_U\subseteq I$ of "relevant coordinates" for $V$ such that for any $y\in X$ which agrees with $x$ on the coordinates in $J_U$, $y\in V\subseteq U$.
Let $I' = \bigcup_{U\in \mathcal{B}_x} J_U$. By counting, $|I'|<|I|$ (this is where we use the fact that $I$ is infinite). So we can pick some $x'\in X$ such that $x'\neq x$, but $x'$ only differs from $x$ on coordinates in $I\setminus I'$. By construction, even open set containing $x$ also contains $x'$. But it is easy to find a standard basic open set $V$ with $x\in V$ and $x'\notin V$, so we have a contradiction.