Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 3 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.

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.