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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?

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up vote 2 down vote accepted

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.

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Excellent; that's a very short and nice proof, thanks. –  Tom Jonathan Jun 14 '12 at 9:21
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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

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