# Infimum of a union

I have a set $X$ and a function $$f: X \rightarrow \mathbb{R}$$ and I am interested in the value $$\inf\limits_{x \in X} f(x) \,.$$ I can represent $X$ as $$X = \bigcup\limits_{i \in I} X_i \,,$$ where the index set $I$ is uncountable. Now I wonder whether $$\inf\limits_{x \in X} f(x) = \inf\limits_{i \in I} \left( \inf\limits_{x \in X_i} f(x) \right) \,.$$ Is this true? If so, how can I see this?

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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.
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