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I have encountered a relation that I couldn't really get my head around it and understand what it does.

The relation is on the set of partial functions from a set S to a set T:

$f \le g \Leftrightarrow \operatorname{dom}(f) \subseteq \operatorname{dom}(g)$, and $s \in \operatorname{dom}(f)$ implies that $fs=gs$

I'm not sure but I think this relates the same partial function

and what are the maximal and minimal elements of this poset? Is it the same function?

Thank you

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2 Answers 2

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Recall that a partial function $f: S\nrightarrow T$ is a function such that $\operatorname{dom}(f)\subseteq S$, whereas typical functions take $S$ to be domain. Onto your poset:

Let $A$ be your set of partial functions of the above form, and define a partial ordering on $A$ in the following way.

For any two partial functions $f,g\in A$ such that

  • $\operatorname{dom}(f) \subseteq \operatorname{dom}(g)$, and
  • For all $s\in\operatorname{dom}(f)$, $f(s) = g(s)$

we say that $f\le g$.

You can think of functions related in this way as subfunctions, where if $f\le g$, then $f$ is the same function as $g$, but you're just mapping fewer elements from $S$ over to $T$. Therefore, a minimal element of $A$ would be a function that maps the fewest elements -- namely, the function that maps nothing! Note that this is also the least element in your poset.

The maximal elements are similar, but not exactly analagous. Note that a partial function may not be a function if its domain were extend to all of $S$ (e.g. the function $f:\mathbb{R}\to\mathbb{R}$ defined by $x\mapsto\frac{1}{x}$). Therefore, the maximal elements are partial functions $f:\operatorname{dom}(f)\to T$ such that for any $s\in S$, $f:(\operatorname{dom}(f)\cup\{s\})\to T$ fails to be a function.

Edit: As @CiaPan's excellent answer points out, there are no largest elements in the case when $S$ and $T$ have more than a single element.

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  • $\begingroup$ I understood the relation now thanks, the minimal and maximal elements are also the least and the greatest since they are the only minimal and maximal, right? Thanks $\endgroup$ Commented Mar 21, 2017 at 12:06
  • $\begingroup$ That's a good point that I overlooked, thank you! I edited my response to reflect the change. $\endgroup$ Commented Mar 21, 2017 at 12:37
  • $\begingroup$ For the example you wrote it is the case iff your function was 1/x, otherwise the greatest element is the element that maps the whole source set, right? Also the greatest lower bound of f and g will be defined by the intersection of Dom(f) and Dom(g), and for the least upper bound it will be defined by their union of the domains, does that sound right? Thanks. $\endgroup$ Commented Mar 21, 2017 at 12:53
  • $\begingroup$ By greatest lower bound, do you mean the greatest function which is less than both $f$ and $g$? $\endgroup$ Commented Mar 21, 2017 at 12:58
  • $\begingroup$ Yeah, but that's not the only condition, like if we take all functions greater than f and g the least upper bound is the function that is relates to both f and g but it's immediately above it (if we looked at the hassle diagram of the relation) $\endgroup$ Commented Mar 21, 2017 at 13:02
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I would say $f\le g$ if $g$ is an extension of $f$ (the domain of $g$ contains the whole domain of $f$ and they are equal on that common part of domains). Then a maximal element is each function which has no extensions and a minimal function is each function which is not an extension of another one.

So each function defined on the whole $S$ is maximal, and each function defined on a single point of $S$ is minimal.

Edit

I'm not quite sure if the empty function $\{\}\to\{\}$ should be considered as a partial function from arbitrary $S$ to any $T$. If so (as @JazzyMatrix assumes in the answer), then of course that will be the least, and obviously the only minimal, element of the poset.

The special cases are:

  • $S$ is empty – depending on the decision on including the empty function, there is either none or there is exactly one function to any $T$, namely the empty function, which is therefore the least and the greatest element in the poset.
  • $S$ is non-empty and $T$ is a singleton (has one element) – each function represents exactly one subset of $S$ (namely its own domain), hence the poset reflects the partial ordering of $P(S)$ by set inclusion. The least element is the empty function (representing the empty subset of $S$), if we include the empty function, or else each function on a single-point domain is a minimal element of the poset. The maximal element is the function $S\to T$.
    • $S$ and $T$ are singletons – there is only one non-empty function and (possibly) the empty function, so we have either a single-item poset or two-items well-ordered set.
  • $S$ is non-empty and $T$ has more than one element – the general case which I described before, but it should be modified now by adding (or not) the empty function, being the least element of the poset.
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