blocks|key|2979688|text|Hask对象是Haskell类型，其数量是无限的。Hask箭头是Haskell函数，也是无限的。因此Hask不仅局部小，Hask也小。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2979689|card(ob(Hask))=card(hom(Hask))=card(N)|2979690|有关Hask的更多详细信息，请单击此处：|2979691|http://yannesposito.com/Scratch/en/blog/Category-Theory-Presentation/|offset|length|2979692|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|0|1X|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|T|8|@]|9|@$H|U|I|V|1|W]]|A|$]]|$1|J|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|G]]]]

Hask objects are Haskell types which are countably infinite. Hask arrows are Haskell functions which are also countably infinite. Therefore Hask is not only locally small, Hask is small. 

card(ob(Hask))=card(hom(Hask))=card(N)

More details about Hask here:

<a href="http://yannesposito.com/Scratch/en/blog/Category-Theory-Presentation/" rel="nofollow">http://yannesposito.com/Scratch/en/blog/Category-Theory-Presentation/</a>

blocks|key|2979723|text|什么是Hask？如果它包含所有haskell可定义的“函数”作为态射，那么肯定不是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2979724|data+Big+=+Big+(Big+->+Big)|code-block|syntax|javascript|2979725|Big+->+Big的"hom+set“包含了整个非类型化的lambda演算！我怀疑它是局部小的，即使你只允许终止函数--我认为system-f没有固定的理论模型。|offset|length|style|CODE|2979726|编辑:七年后，我无法理解我在这里想要说的话。Hask没有集合理论模型，因为模型将函数类型解释为完整的函数集。这是真的，但我不知道这和这个问题有什么关系。不太清楚"Hask“是什么，但在我看来，任何合理的答案都会有小的homsets+(也就是说，它是局部小的)。|2979727|我多年前回答的奇怪之处让我有点尴尬。我确信我的意思是非常有洞察力的--我只是不知道那是什么，而且从字面上看，它似乎是相当错误的。|2979728|entityMap^0|0|0|0|A|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

What is Hask? If it includes all the haskell definable &quot;functions&quot; as morphism then definitly not
<pre><code>data Big = Big (Big -&gt; Big)
</code></pre>
the &quot;hom set&quot; of <code>Big -&gt; Big</code> contains the entire untyped lambda calculus! I doubt it is locally small even if you only allow terminating functions--I think there are no set theoretic models of system-f.
EDIT: seven years later I can't make heads or tails of what I was trying to say here. Hask has no set theoretic models, in the sense of models which interpret function types as the full sets of functions. That is true, but I don't know what that has to do with the question. It isn't really clear what &quot;Hask&quot; is, but any reasonable answer seems to me would have small homsets (that is, it is locally small).
The strangeness of my answer from many years past is slightly embarrassing to me. I'm sure I meant something very insightful--I just have no idea what that was, and as worded it seems rather wrong.

blocks|key|19086|text|尤其是@PhillipJF，这是一个尝试。我并不是想做Hask最精确或最优雅的模型，我只是想做一个模型。请给我评论。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|19087|如果A是Haskell类型，则在Hask中将类型A的值定义为类型A的良好类型Haskell术语的等价类(类型检查器将接受x+::+A的字符串x)，模扩展相等。也就是说，如果两个项扩展到相同的(可能是无限的)范式，那么它们被认为是相等的，并且两个没有hnf的项也是相等的。事实上，这是不可决定的，这是无关紧要的，我们只需要说明这些条件集-理论上，我几乎怀疑我们可以做到。|offset|length|style|CODE|19088|假设Haskell+Hask的对象是Haskell类型(原始类型和用户定义的类型；我们将假设所有用户定义的类型都存在并具有不同的名称。用户定义的类型定义是源代码，因此它们是可计数的。把它们命名为D0，D1，..。根据该计数。)。|19089|设态射A_+->+_B为A_型->+_B的值|19090|设A上的恒等式是id+::+A+->+A的等价类，同样地，设g和f的合成是g+.+f的等价类。|19091|所有值的集合是一个可数集合，因为术语只是有限字母表上的字符串。所以这个Hask的模型很小。|BOLD|19092|这是错的吗？|19093|entityMap^0|0|1O|6|0|2P|2|2S|2|0|0|8|C|U|1|W|1|11|5|0|Z|4|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@$D|W|E|X|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|Y|8|@$D|Z|E|10|F|G]|$D|11|E|12|F|G]]|9|@]|A|$]]|$1|J|3|K|5|6|7|13|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|14|8|@$D|15|E|16|F|G]|$D|17|E|18|F|G]|$D|19|E|1A|F|G]|$D|1B|E|1C|F|G]]|9|@]|A|$]]|$1|N|3|O|5|6|7|1D|8|@$D|1E|E|1F|F|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1G|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|T|$]]

Especially @PhillipJF, here's a try. I'm not trying to make the most accurate or elegant model of Hask, I'm just trying to make a model. Critique, please.

If A is a Haskell type, define a value of type A in Hask to be an equivalence class of well-typed Haskell terms of type A (strings x for which <code>x :: A</code> would be accepted by the type checker), modulo extensional equality. That is, two terms are considered equal if they expand out to the same (possibly infinite) normal form, and two terms which have no hnf are also equal. The fact that this is not decidable is irrelevant, we need only state these conditions set-theoretically, which I have little doubt we can do.

Let the objects of Hask be Haskell types (primitive types &amp; user-defined types; we will assume that all user-definable types exist and have distinct names. User-defined type definitions are source code, so they are countable. Just name them <code>D0</code>,<code>D1</code>, ... according to that counting.).

Let the morphisms A -> B be values of type A -> B

Let the identity on A be the equivalence class of <code>id :: A -&gt; A</code>, and similarly let composition of <code>g</code> and <code>f</code> be the equivalence class of <code>g . f</code>.

The set of all values is a countable set, because terms are just strings over a finite alphabet. So this model of Hask is small.

Is this wrong?

Is the category Hask of haskell objects an example of a locally small category? 

<a href="http://ncatlab.org/nlab/show/locally+small+category" rel="nofollow">http://ncatlab.org/nlab/show/locally+small+category</a>

Maybe not.. hask as cpo
<a href="http://www.cs.gunma-u.ac.jp/~hamana/Papers/cpo.pdf" rel="nofollow">http://www.cs.gunma-u.ac.jp/~hamana/Papers/cpo.pdf</a>

The haskellwiki, <a href="http://www.haskell.org/haskellwiki/Hask" rel="nofollow">http://www.haskell.org/haskellwiki/Hask</a> has very good information, showing that Hask is not Cartesian Closed.

Is Hask locally small?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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haskell对象的Hask类别是局部小类别的一个例子吗？也许不是..伪装成首席执行官haskellwiki，有非常好的信息，说明Hask不是笛卡尔闭合的。

问Hask在本地是小的吗？
EN

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问Hask在本地是小的吗？EN