Steven D'Aprano wrote:
> On Sun, Mar 22, 2020 at 07:59:59PM -0000, Steve Jorgensen wrote:
> > Currently, the issubset and
> > issuperset methods of set objects
> > accept arbitrary iterables as arguments. An iterable that is both a
> > subset and superset is, in a sense, "equal" to the set. It would be
> > inappropriate for == to return True for such a comparison,
> > however, since that would break the Hashable contract.
> > I think the "arbitrary iterables" part is a distraction. We are
> fundamentally talking about a comparison on sets, even if Python relaxes
> the requirements and also allows one operand to be a arbitrary iterable.
> I don't believe that a set A can be both a superset and subset of
> another set B at the same time. On a Venn Diagram, that would require A
> to be both completely surrounded by B and B to be completely surrounded
> by A at the same time, which is impossible.
> I think you might be talking about sets which partially overlap:
> A = {1, 2, 3, 4}
> B = {2, 3, 4, 5}
Every set is a superset of itself and a subset of itself. A set may not be a
"formal" subset or a "formal" superset of itself. `issubset` and `issuperset`
refer to standard subsets and supersets, not formal subsets and supersets.
In Python, you can trivially check that…
```
In [1]: {1, 2, 3}.issubset({1, 2, 3})
Out[1]: True
In [2]: {1, 2, 3}.issuperset({1, 2, 3})
Out[2]: True
In [3]: {1, 2, 3}.issubset((1, 2, 3))
Out[3]: True
In [4]: {1, 2, 3}.issuperset((1, 2, 3))
Out[4]: True
```
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