Infimum of a union I have a set $X$ and a function
\begin{equation}
f: X \rightarrow \mathbb{R}
\end{equation}
and I am interested in the value
\begin{equation}
\inf\limits_{x \in X} f(x) \,.
\end{equation}
I can represent $X$ as
\begin{equation}
X = \bigcup\limits_{i \in I} X_i \,,
\end{equation}
where the index set $I$ is uncountable. Now I wonder whether
\begin{equation}
\inf\limits_{x \in X} f(x) = \inf\limits_{i \in I} \left( \inf\limits_{x \in X_i} f(x) \right) \,.
\end{equation}
Is this true? If so, how can I see this?
 A: Yes, it’s true. Let $\alpha=\inf\limits_{x\in X}f(x)$, and for each $i\in I$ let $\alpha_i=\inf\limits_{x\in X_i}f(x)$. Clearly $\alpha\le\alpha_i$ for each $i\in I$, so $\alpha\le\inf\limits_{i\in I}\alpha_i$. 
On the other hand, for each $x\in X$ there is an $i\in I$ such that $x\in X_i$ and therefore $\alpha_i\le f(x)$, so $\inf\limits_{i\in I}\alpha_i\le\inf\limits_{x\in X}f(x)=\alpha$. (Alternatively if $\alpha<\inf\limits_{i\in I}\alpha_i$, then there is some $x\in X$ such that $$\alpha\le f(x)<\inf_{i\in I}\alpha_i\;.$$ But then $f(x)<\alpha_i$ for all $i\in I$, which is clearly impossible.)
Note that the cardinality of $I$ doesn’t matter.
A: by the infimum property there a $x_0 \in X$ such that 
$$
f(x_0) \leq \inf_{x \in X}f(x) +\epsilon
$$
but $ x_0 \in X$ so there a $X_k$ such that $ x_0 \in X_k$
$$ f(x_0) \geq \inf_{x \in X_k}f(x)$$ therefore
$$f(x_0) \geq \inf_{i \in I} \inf_{x \in X_i}f(x)$$
so $$
\inf_{i \in I} \inf_{x \in X_i}f(x) \leq \inf_{x \in X}f(x) +\epsilon
$$ 
Again by the infimum property choose an $\lambda$ such that
$$\inf_{x \in X_{\lambda}}f(x)  \leq \inf_{i \in I} \inf_{x \in X_i}f(x) +\epsilon$$
hence $$\inf_{x \in X_{\lambda}}f(x) \geq \inf_{x \in X} f(x)$$ because 
$X_{\lambda} \subset X$ combining the last two the opposite direction is proved so you are done
