Infimum of a proper subset contained in a set with a lower bound.

Question

If I have some set $A$ = {$ f(x) | x \in \mathbb{F}$}

Where $f: \mathbb{F} \Rightarrow \mathbb{L}$

Then $A \subset \mathbb{L}$

Now if Inf ($\mathbb{L})=\alpha$

And $\alpha \in A$

Then Inf($A$) = $\alpha$

Is this always true?

Can it be used to prove a value of Inf($A$) if there is a defined value of Inf($\mathbb{L}$) or do we have to assume it is just a lower bound of $A$ and continue to prove that it is Inf($A$) through the standard conditions of an infimum.

Answer 1

Yes it's always true.

It follows from that a lower bound that is member of the set is the infimum - it follows from the definition: if a lower bound $\alpha$ of $A$ it's obviously a lower bound the question is if it's the largest - assume it wasn't, that there is a $\beta>\alpha$ that is a lower bound, but we have $\alpha\in A$ and $\alpha<\beta$ so $\beta$ is not a lower bound of $A$.

You also use that a lower bound of a superset is a lower bound to the set itself. That is if $\alpha$ is a lower bound (infimum is a lower bound) of $\mathbb L$ we have that $\alpha\le x$ for all $x\in\mathbb L$. Now if $A\subseteq\mathbb L$ we have that any $x\in A$ we have that $x\in\mathbb L$ and therefore $\alpha\le x$ - that is $\alpha$ is a lower bound of $A$.

Note that this technique can only be used to find the infimum of $A$ if it actually contains the infimum of $\mathbb L$.

Answer 2

Edit:

To address your revisions. It is not guaranteed that $\inf(L)$ can be contained in the set $A$. For instance, consider a map into $\mathbb{R}$. $Inf(\mathbb{R})$ is typically taken to be $-\infty \notin \mathbb{R}$.Thus as $A \subseteq \mathbb{R}$, we cannot have $inf(\mathbb{R}) \in A$.

But if that is possible, then yes we can deduce that $inf(L) = inf(A)$. $inf(A)$ is defined to be the greatest lower bound of $A$. Thus is $inf(L)$ exists and is contained in $A$, there can be no smaller element in $A$ so we have $inf(L) = inf(A)$.

Old answer:

Let $A$ be defined as above and $\mathbb{R}^{+} = [0, +\infty)$. Then it is clear that $A = \mathbb{R}^{+}$. As $A$ is bounded below inclusive, then $inf(A) = min(A) = 0$, as Adren noted. So yes, $inf(A) = inf(\mathbb{R}^{+})$. More generally, check out the Extreme Value theorem to learn more about when continous functions attain their extreme values. Also as a note, $\Re$ is nonstandard notation for the real numbers, and $\Re^{+}$ even more so. Generally $\mathbb{R}$ is used, which is \mathbb{R} in latex.