How does one extend the notion of infimum to Metric Spaces?

Question

In particular I was trying to make sense of the expression:

$$ dist(V,f) = \inf_{v \in V} \| v - f \|_{\mathcal M}$$

which seems to require me to extend the notion of infimum to metric spaces and things that look like minimization problems.

To start this search I reviewed the definition of infimum from Walter Rudin's book (here paraphrased):

Suppose $(P, \leq)$ is an ordered set, $S \subset P$ and $S$ is bounded from bellow. Suppose there exists an $\alpha \in P$ with the following properties:

1. $\alpha$ is a lower bound. i.e. $\forall x \in S, \alpha \leq x$. i.e. $\alpha \in LowerBounds(S)$
2. If $\gamma > \alpha$ then $\gamma \in P$ is not a lower bound. i.e. if $\gamma > \alpha \implies \gamma \notin LowerBounds(S)$.

which makes sense and the notation:

$$ \alpha = \inf S$$

makes sense to me. Similarly I think:

$$ \alpha = \inf_{ x \in P} S $$

makes sense to me too, just being more explicit over which set (i.e. $P$) is the infimum and with respect to what set (i.e. $S$) we are taking the infimum.

However, I am having difficulties extending this notation because in a metric space we don't explicitly have an ordered set nor do we have the explicit sets $P$ and $S$ defined.

It seems that for this to make sense or be equivalent we would have to define $P = \mathbb R$ the reals and $S_f = \{ \| v - f \|_{\mathcal M} : f \in V\}$ which seems in this case that $S_f \subset \mathbb R$ and the ordered set is induced by the standard order on the number line. The issue in this case might be that multiple different $v \in V$ could map to the same real number depending on the nature of $V$. So in this case it seems that the infimum would be the smallest numerical/real value for $\| v - f \|_{\mathcal M}$ and that specific number is unique but it seems multiple different $v \in V$ could achieve it (at least in principle). So the value achieving the infimum seems to not be unique.

However, something else worries me. It worries me that each value of $v \in V$ decreases $\| v - f \|_{\mathcal M}$ a little bit more and more, making it a sequence that decreases forever and that approaches some value (say $L_{B_f}$). If this is true it seems that the infimum wouldn't exist, at least not using element from $V$, right? Would I have to change $V$ in this case? Do I really need the word infimum if its not achievable? I know that sometimes one cannot achieve the minimum so we use the word infimum. Is this one of those examples?

To make it clear my main worries are two:

1. What the role of $\| v - f\|_{\mathcal M}$ and $V$ is. It seems that $\| v - f\|_{\mathcal M}$ is just for us to define the set we are looking to define a lower bound and that $V$ is just an index set.
2. It seems that infimum has a small assumption that I just realized (that isn't clear until you "view it" as an optimization problem. In this case $\alpha = \inf S \iff \alpha = \inf_{ x \in P} S $ which is important to make explicit what "the minimization is over". It seems that the minimization is always over the set $P$ i.e. the larger set that $S$ is a susbet of. But if the minimization was over something else that did not have say, a largest element itself and we had: $ \inf_{x \in Q} S$ where we are looking to lower bound $S$ with $Q$ but $Q$ itself might a subset of $P$ but not have a largest element. In this case the "infimum" would not be defined. Maybe this is not what the definition of infimum is and it should be that we look for a lowest value that actually exists. Otherwise, there is largest lower bound...right?

Answer 1

Actually, you don't need to be concerned about the metric space at all. The metric space only serves as the index set for the collection of real numbers $|| v-f||_{\mathcal{M}}$.

Answer 2

You seem to be confused about the notation. The infimum is defined as the largest lower bound of an ordered set (in your example $P$), usually a subset of the real numbers. It always exists if $P$ is not empty (possibly it is -$\infty$).

The expression

$\alpha = \inf_{x\in P} S$

does not make sense. In general yo use the notation $\inf S$. However, if the set $S$ is given by the image of some function $f:X \to\mathbb{R}$, it is convenient to write $$\inf_{x \in X} f(x),$$ which stands for $$\inf \{f(x)|~{x \in X}\}=\inf f(X).$$

EDIT: I dont have enough reputation to comment, so i write it here...

I didn't write any definition, so i am not quite sure what you are refering to. There are some equivalent definitions of the infimum, one of them is, that the infimum of $S\subset\mathbb{R}$ is the greatest lower bound of $S$. It always exists, because $-\infty$ is a lower bound to every non empty subset of $\mathbb{R}$. And as there exists a lower bound, there also exists a greatest lower bound. Note that this does not mean that $-\infty\in S$ (which doesn't make sense anyway).

Answer 3

Let $X $ be a metric space with $||x-y|| ;x,y\in X $ being the notation for the metric "distance" between $x $ and $y $.

Let $V\subset X $. Let $f\in X $. $f $ may or may not be in $V $ but in general we should assume it doesn't.

Let $S \subset R ; S =\{||f-v||:v \in V\} $. In other words, $S $ is the set of distances from $f $ to the various points of $V $. Now for any $d\in S $, $d \ge 0$ and if $f \not \in V $ then $d=||f-v||$ for some $v \in V $ so $d > 0$.

So $S $ is bounded below by $0$. So assuming $V $ is non empty, $S $ is non empty so $\inf S $ exists and $\inf S \ge 0$.

Define $dist(f,V) =\inf S$. Notationwise, as this value depends on $f $ and the set $V $ we can notate this as $dist (f,V)= \inf_{v\in V}||f-v||=\inf\{||f-v||:v\in V\} $.

Now it may be possible if $f \not\in V $, that for all $v \in V $ that $||f-v|| > dist(f,V) $. In other words there is no single nearest point of $V $ to $f $. It's possible if $f $ is a limit point not in $V $ that $dist (f,V)=0$.

But if $f $ is not in $V $ and $f $ is not a limit point of $V $ then there is an $r>0$ so that $B (f,r) $ has no point of $V $ so for all $v \in V$, $||f-v|| \ge r $ so $dist(f,V) \ge r > 0$.