When is it okay to intersect infinite families of proper classes?












3












$begingroup$


For experts who work in ZFC, it is common knowledge that one cannot in general define a countable intersection/union of proper classes. However, in my work as a ring theorist I intersect infinite collections of proper classes all the time.



Here is a simple example. Let a ring $R$ be called $n$-Dedekind-finite if $ab=1implies ba=1$ for $a,bin mathbb{M}_n(R)$ (the $ntimes n$ matrix ring over $R$). A ring $R$ is said to be stably finite if it is $n$-Dedekind-finite for every integer $ngeq 1$. The class of stably finite rings is the intersection (over positive integers $n$) of the classes of $n$-Dedekind-finite rings.



So my question is a straightforward one. What principle makes it okay for me to intersect classes in this manner?



Phrased a little differently my question is the following. Suppose we have an $mathbb{N}$-indexed collection of sentences $S_n$ in the first-order language of rings. Why is it valid (in my day-to-day work) to form the class of rings satisfying $land_{nin mathbb{N}}S_n$, even though the corresponding class doesn't necessarily exist if each $S_n$ is a sentence in the language of ZFC?



[Part of my motivation for this question is the idea that much of "normal mathematics" can be done in ZFC. That's how I've viewed my own work. I'm happy to think of my rings living in $V$, subject to all the constraints of ZFC. (In some cases I might go further, by invoking the existence of universes or the continuum hypothesis, but that is not the norm.) But I'm unsure how to justify my use of infinite Boolean operations on proper classes. Especially since, in some cases, my conditions on the rings are conditions on sets, from the language of ZFC!]






set-theory lo.logic ra.rings-and-algebras







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    This is because here you don't have indepently generated sentences $S_n$, you have a single formula $varphi(n,R)$ and so you can simply define the formula $forall n, varphi(n, R)$.
    $endgroup$
    – Max
    51 mins ago












  • $begingroup$
    @Max See my response to Andrej below.
    $endgroup$
    – Pace Nielsen
    37 mins ago






  • 1




    $begingroup$
    I don't get your point. Why would $n$ be a "meta-integer" ? What would $M_n(R)$ even mean if it were ?
    $endgroup$
    – Max
    27 mins ago










  • $begingroup$
    @Max The axioms of ZFC are countable, and you can list them as $A_1,A_2,ldots$. The subscripts are not elements of the set $mathbb{N}$ inside ZFC, but rather are from the meta-theory. You cannot collect all of the axioms into the single statement $forall n A_n$, because the $n$'s are not ranging over a set in ZFC.
    $endgroup$
    – Pace Nielsen
    17 mins ago






  • 1




    $begingroup$
    The definition of an $n$-DF ring is internal to the universe of sets. Not external.
    $endgroup$
    – Asaf Karagila
    8 mins ago
















3












$begingroup$


For experts who work in ZFC, it is common knowledge that one cannot in general define a countable intersection/union of proper classes. However, in my work as a ring theorist I intersect infinite collections of proper classes all the time.



Here is a simple example. Let a ring $R$ be called $n$-Dedekind-finite if $ab=1implies ba=1$ for $a,bin mathbb{M}_n(R)$ (the $ntimes n$ matrix ring over $R$). A ring $R$ is said to be stably finite if it is $n$-Dedekind-finite for every integer $ngeq 1$. The class of stably finite rings is the intersection (over positive integers $n$) of the classes of $n$-Dedekind-finite rings.



So my question is a straightforward one. What principle makes it okay for me to intersect classes in this manner?



Phrased a little differently my question is the following. Suppose we have an $mathbb{N}$-indexed collection of sentences $S_n$ in the first-order language of rings. Why is it valid (in my day-to-day work) to form the class of rings satisfying $land_{nin mathbb{N}}S_n$, even though the corresponding class doesn't necessarily exist if each $S_n$ is a sentence in the language of ZFC?



[Part of my motivation for this question is the idea that much of "normal mathematics" can be done in ZFC. That's how I've viewed my own work. I'm happy to think of my rings living in $V$, subject to all the constraints of ZFC. (In some cases I might go further, by invoking the existence of universes or the continuum hypothesis, but that is not the norm.) But I'm unsure how to justify my use of infinite Boolean operations on proper classes. Especially since, in some cases, my conditions on the rings are conditions on sets, from the language of ZFC!]






set-theory lo.logic ra.rings-and-algebras







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    This is because here you don't have indepently generated sentences $S_n$, you have a single formula $varphi(n,R)$ and so you can simply define the formula $forall n, varphi(n, R)$.
    $endgroup$
    – Max
    51 mins ago












  • $begingroup$
    @Max See my response to Andrej below.
    $endgroup$
    – Pace Nielsen
    37 mins ago






  • 1




    $begingroup$
    I don't get your point. Why would $n$ be a "meta-integer" ? What would $M_n(R)$ even mean if it were ?
    $endgroup$
    – Max
    27 mins ago










  • $begingroup$
    @Max The axioms of ZFC are countable, and you can list them as $A_1,A_2,ldots$. The subscripts are not elements of the set $mathbb{N}$ inside ZFC, but rather are from the meta-theory. You cannot collect all of the axioms into the single statement $forall n A_n$, because the $n$'s are not ranging over a set in ZFC.
    $endgroup$
    – Pace Nielsen
    17 mins ago






  • 1




    $begingroup$
    The definition of an $n$-DF ring is internal to the universe of sets. Not external.
    $endgroup$
    – Asaf Karagila
    8 mins ago














3












3








3





$begingroup$


For experts who work in ZFC, it is common knowledge that one cannot in general define a countable intersection/union of proper classes. However, in my work as a ring theorist I intersect infinite collections of proper classes all the time.



Here is a simple example. Let a ring $R$ be called $n$-Dedekind-finite if $ab=1implies ba=1$ for $a,bin mathbb{M}_n(R)$ (the $ntimes n$ matrix ring over $R$). A ring $R$ is said to be stably finite if it is $n$-Dedekind-finite for every integer $ngeq 1$. The class of stably finite rings is the intersection (over positive integers $n$) of the classes of $n$-Dedekind-finite rings.



So my question is a straightforward one. What principle makes it okay for me to intersect classes in this manner?



Phrased a little differently my question is the following. Suppose we have an $mathbb{N}$-indexed collection of sentences $S_n$ in the first-order language of rings. Why is it valid (in my day-to-day work) to form the class of rings satisfying $land_{nin mathbb{N}}S_n$, even though the corresponding class doesn't necessarily exist if each $S_n$ is a sentence in the language of ZFC?



[Part of my motivation for this question is the idea that much of "normal mathematics" can be done in ZFC. That's how I've viewed my own work. I'm happy to think of my rings living in $V$, subject to all the constraints of ZFC. (In some cases I might go further, by invoking the existence of universes or the continuum hypothesis, but that is not the norm.) But I'm unsure how to justify my use of infinite Boolean operations on proper classes. Especially since, in some cases, my conditions on the rings are conditions on sets, from the language of ZFC!]






set-theory lo.logic ra.rings-and-algebras







share|cite|improve this question









$endgroup$




For experts who work in ZFC, it is common knowledge that one cannot in general define a countable intersection/union of proper classes. However, in my work as a ring theorist I intersect infinite collections of proper classes all the time.



Here is a simple example. Let a ring $R$ be called $n$-Dedekind-finite if $ab=1implies ba=1$ for $a,bin mathbb{M}_n(R)$ (the $ntimes n$ matrix ring over $R$). A ring $R$ is said to be stably finite if it is $n$-Dedekind-finite for every integer $ngeq 1$. The class of stably finite rings is the intersection (over positive integers $n$) of the classes of $n$-Dedekind-finite rings.



So my question is a straightforward one. What principle makes it okay for me to intersect classes in this manner?



Phrased a little differently my question is the following. Suppose we have an $mathbb{N}$-indexed collection of sentences $S_n$ in the first-order language of rings. Why is it valid (in my day-to-day work) to form the class of rings satisfying $land_{nin mathbb{N}}S_n$, even though the corresponding class doesn't necessarily exist if each $S_n$ is a sentence in the language of ZFC?



[Part of my motivation for this question is the idea that much of "normal mathematics" can be done in ZFC. That's how I've viewed my own work. I'm happy to think of my rings living in $V$, subject to all the constraints of ZFC. (In some cases I might go further, by invoking the existence of universes or the continuum hypothesis, but that is not the norm.) But I'm unsure how to justify my use of infinite Boolean operations on proper classes. Especially since, in some cases, my conditions on the rings are conditions on sets, from the language of ZFC!]






set-theory lo.logic ra.rings-and-algebras




set-theory lo.logic ra.rings-and-algebras






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









Pace NielsenPace Nielsen

7,15223175




7,15223175








  • 2




    $begingroup$
    This is because here you don't have indepently generated sentences $S_n$, you have a single formula $varphi(n,R)$ and so you can simply define the formula $forall n, varphi(n, R)$.
    $endgroup$
    – Max
    51 mins ago












  • $begingroup$
    @Max See my response to Andrej below.
    $endgroup$
    – Pace Nielsen
    37 mins ago






  • 1




    $begingroup$
    I don't get your point. Why would $n$ be a "meta-integer" ? What would $M_n(R)$ even mean if it were ?
    $endgroup$
    – Max
    27 mins ago










  • $begingroup$
    @Max The axioms of ZFC are countable, and you can list them as $A_1,A_2,ldots$. The subscripts are not elements of the set $mathbb{N}$ inside ZFC, but rather are from the meta-theory. You cannot collect all of the axioms into the single statement $forall n A_n$, because the $n$'s are not ranging over a set in ZFC.
    $endgroup$
    – Pace Nielsen
    17 mins ago






  • 1




    $begingroup$
    The definition of an $n$-DF ring is internal to the universe of sets. Not external.
    $endgroup$
    – Asaf Karagila
    8 mins ago














  • 2




    $begingroup$
    This is because here you don't have indepently generated sentences $S_n$, you have a single formula $varphi(n,R)$ and so you can simply define the formula $forall n, varphi(n, R)$.
    $endgroup$
    – Max
    51 mins ago












  • $begingroup$
    @Max See my response to Andrej below.
    $endgroup$
    – Pace Nielsen
    37 mins ago






  • 1




    $begingroup$
    I don't get your point. Why would $n$ be a "meta-integer" ? What would $M_n(R)$ even mean if it were ?
    $endgroup$
    – Max
    27 mins ago










  • $begingroup$
    @Max The axioms of ZFC are countable, and you can list them as $A_1,A_2,ldots$. The subscripts are not elements of the set $mathbb{N}$ inside ZFC, but rather are from the meta-theory. You cannot collect all of the axioms into the single statement $forall n A_n$, because the $n$'s are not ranging over a set in ZFC.
    $endgroup$
    – Pace Nielsen
    17 mins ago






  • 1




    $begingroup$
    The definition of an $n$-DF ring is internal to the universe of sets. Not external.
    $endgroup$
    – Asaf Karagila
    8 mins ago








2




2




$begingroup$
This is because here you don't have indepently generated sentences $S_n$, you have a single formula $varphi(n,R)$ and so you can simply define the formula $forall n, varphi(n, R)$.
$endgroup$
– Max
51 mins ago






$begingroup$
This is because here you don't have indepently generated sentences $S_n$, you have a single formula $varphi(n,R)$ and so you can simply define the formula $forall n, varphi(n, R)$.
$endgroup$
– Max
51 mins ago














$begingroup$
@Max See my response to Andrej below.
$endgroup$
– Pace Nielsen
37 mins ago




$begingroup$
@Max See my response to Andrej below.
$endgroup$
– Pace Nielsen
37 mins ago




1




1




$begingroup$
I don't get your point. Why would $n$ be a "meta-integer" ? What would $M_n(R)$ even mean if it were ?
$endgroup$
– Max
27 mins ago




$begingroup$
I don't get your point. Why would $n$ be a "meta-integer" ? What would $M_n(R)$ even mean if it were ?
$endgroup$
– Max
27 mins ago












$begingroup$
@Max The axioms of ZFC are countable, and you can list them as $A_1,A_2,ldots$. The subscripts are not elements of the set $mathbb{N}$ inside ZFC, but rather are from the meta-theory. You cannot collect all of the axioms into the single statement $forall n A_n$, because the $n$'s are not ranging over a set in ZFC.
$endgroup$
– Pace Nielsen
17 mins ago




$begingroup$
@Max The axioms of ZFC are countable, and you can list them as $A_1,A_2,ldots$. The subscripts are not elements of the set $mathbb{N}$ inside ZFC, but rather are from the meta-theory. You cannot collect all of the axioms into the single statement $forall n A_n$, because the $n$'s are not ranging over a set in ZFC.
$endgroup$
– Pace Nielsen
17 mins ago




1




1




$begingroup$
The definition of an $n$-DF ring is internal to the universe of sets. Not external.
$endgroup$
– Asaf Karagila
8 mins ago




$begingroup$
The definition of an $n$-DF ring is internal to the universe of sets. Not external.
$endgroup$
– Asaf Karagila
8 mins ago










1 Answer
1






active

oldest

votes


















5












$begingroup$

You will have to come up with trickier examples.



There is a formula $phi(R, n)$ of ZFC with parameters $R$ and $n$ whose meaning is "$R$ is an $n$-Dedekind finite ring". I hope this much is clear (and true!). The class of all stably finite rings is simply ${R mid forall n geq 1 ,., phi(R, n)}$. In other words, the imagined intersection of a countable collection of classes is not needed at all, because
$$bigcap_{n geq 1} {R mid phi(R, n)} = {R mid forall n geq 1 ,., phi(R, n)}.$$
The point is that the property of being $n$-Dedekind finite is uniform in $n$ and so a single formula $phi(n, R)$ works. The situation would be different if you proposed an infinite sequence of classes
$$C_1, C_2, C_3, ldots$$
each defined by some formula $psi_i$, so that $C_i = {x mid psi_i(x)}$, such that there is no single formula $Psi(i, x)$ of ZFC for which $psi_i(x) Leftrightarrow Psi(i,x)$. Such examples occur in logic, but I've never seen any "normal" mathematics producing a sequence of highly non-uniform statements.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I don't think it is that easy. The $n$ here is a meta-natural number (not a natural number from inside ZFC, but from my meta-theory). So there is no such formula $phi(R,n)$ of ZFC.
    $endgroup$
    – Pace Nielsen
    42 mins ago










  • $begingroup$
    That said, go ahead and replace the family I gave with a new family that isn't given with any indexing (either from inside or outside the theory). In my day-to-day practice, I'd still think that the intersection of classes exists.
    $endgroup$
    – Pace Nielsen
    38 mins ago











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









5












$begingroup$

You will have to come up with trickier examples.



There is a formula $phi(R, n)$ of ZFC with parameters $R$ and $n$ whose meaning is "$R$ is an $n$-Dedekind finite ring". I hope this much is clear (and true!). The class of all stably finite rings is simply ${R mid forall n geq 1 ,., phi(R, n)}$. In other words, the imagined intersection of a countable collection of classes is not needed at all, because
$$bigcap_{n geq 1} {R mid phi(R, n)} = {R mid forall n geq 1 ,., phi(R, n)}.$$
The point is that the property of being $n$-Dedekind finite is uniform in $n$ and so a single formula $phi(n, R)$ works. The situation would be different if you proposed an infinite sequence of classes
$$C_1, C_2, C_3, ldots$$
each defined by some formula $psi_i$, so that $C_i = {x mid psi_i(x)}$, such that there is no single formula $Psi(i, x)$ of ZFC for which $psi_i(x) Leftrightarrow Psi(i,x)$. Such examples occur in logic, but I've never seen any "normal" mathematics producing a sequence of highly non-uniform statements.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I don't think it is that easy. The $n$ here is a meta-natural number (not a natural number from inside ZFC, but from my meta-theory). So there is no such formula $phi(R,n)$ of ZFC.
    $endgroup$
    – Pace Nielsen
    42 mins ago










  • $begingroup$
    That said, go ahead and replace the family I gave with a new family that isn't given with any indexing (either from inside or outside the theory). In my day-to-day practice, I'd still think that the intersection of classes exists.
    $endgroup$
    – Pace Nielsen
    38 mins ago
















5












$begingroup$

You will have to come up with trickier examples.



There is a formula $phi(R, n)$ of ZFC with parameters $R$ and $n$ whose meaning is "$R$ is an $n$-Dedekind finite ring". I hope this much is clear (and true!). The class of all stably finite rings is simply ${R mid forall n geq 1 ,., phi(R, n)}$. In other words, the imagined intersection of a countable collection of classes is not needed at all, because
$$bigcap_{n geq 1} {R mid phi(R, n)} = {R mid forall n geq 1 ,., phi(R, n)}.$$
The point is that the property of being $n$-Dedekind finite is uniform in $n$ and so a single formula $phi(n, R)$ works. The situation would be different if you proposed an infinite sequence of classes
$$C_1, C_2, C_3, ldots$$
each defined by some formula $psi_i$, so that $C_i = {x mid psi_i(x)}$, such that there is no single formula $Psi(i, x)$ of ZFC for which $psi_i(x) Leftrightarrow Psi(i,x)$. Such examples occur in logic, but I've never seen any "normal" mathematics producing a sequence of highly non-uniform statements.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I don't think it is that easy. The $n$ here is a meta-natural number (not a natural number from inside ZFC, but from my meta-theory). So there is no such formula $phi(R,n)$ of ZFC.
    $endgroup$
    – Pace Nielsen
    42 mins ago










  • $begingroup$
    That said, go ahead and replace the family I gave with a new family that isn't given with any indexing (either from inside or outside the theory). In my day-to-day practice, I'd still think that the intersection of classes exists.
    $endgroup$
    – Pace Nielsen
    38 mins ago














5












5








5





$begingroup$

You will have to come up with trickier examples.



There is a formula $phi(R, n)$ of ZFC with parameters $R$ and $n$ whose meaning is "$R$ is an $n$-Dedekind finite ring". I hope this much is clear (and true!). The class of all stably finite rings is simply ${R mid forall n geq 1 ,., phi(R, n)}$. In other words, the imagined intersection of a countable collection of classes is not needed at all, because
$$bigcap_{n geq 1} {R mid phi(R, n)} = {R mid forall n geq 1 ,., phi(R, n)}.$$
The point is that the property of being $n$-Dedekind finite is uniform in $n$ and so a single formula $phi(n, R)$ works. The situation would be different if you proposed an infinite sequence of classes
$$C_1, C_2, C_3, ldots$$
each defined by some formula $psi_i$, so that $C_i = {x mid psi_i(x)}$, such that there is no single formula $Psi(i, x)$ of ZFC for which $psi_i(x) Leftrightarrow Psi(i,x)$. Such examples occur in logic, but I've never seen any "normal" mathematics producing a sequence of highly non-uniform statements.






share|cite|improve this answer









$endgroup$



You will have to come up with trickier examples.



There is a formula $phi(R, n)$ of ZFC with parameters $R$ and $n$ whose meaning is "$R$ is an $n$-Dedekind finite ring". I hope this much is clear (and true!). The class of all stably finite rings is simply ${R mid forall n geq 1 ,., phi(R, n)}$. In other words, the imagined intersection of a countable collection of classes is not needed at all, because
$$bigcap_{n geq 1} {R mid phi(R, n)} = {R mid forall n geq 1 ,., phi(R, n)}.$$
The point is that the property of being $n$-Dedekind finite is uniform in $n$ and so a single formula $phi(n, R)$ works. The situation would be different if you proposed an infinite sequence of classes
$$C_1, C_2, C_3, ldots$$
each defined by some formula $psi_i$, so that $C_i = {x mid psi_i(x)}$, such that there is no single formula $Psi(i, x)$ of ZFC for which $psi_i(x) Leftrightarrow Psi(i,x)$. Such examples occur in logic, but I've never seen any "normal" mathematics producing a sequence of highly non-uniform statements.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 48 mins ago









Andrej BauerAndrej Bauer

30.3k477166




30.3k477166












  • $begingroup$
    I don't think it is that easy. The $n$ here is a meta-natural number (not a natural number from inside ZFC, but from my meta-theory). So there is no such formula $phi(R,n)$ of ZFC.
    $endgroup$
    – Pace Nielsen
    42 mins ago










  • $begingroup$
    That said, go ahead and replace the family I gave with a new family that isn't given with any indexing (either from inside or outside the theory). In my day-to-day practice, I'd still think that the intersection of classes exists.
    $endgroup$
    – Pace Nielsen
    38 mins ago


















  • $begingroup$
    I don't think it is that easy. The $n$ here is a meta-natural number (not a natural number from inside ZFC, but from my meta-theory). So there is no such formula $phi(R,n)$ of ZFC.
    $endgroup$
    – Pace Nielsen
    42 mins ago










  • $begingroup$
    That said, go ahead and replace the family I gave with a new family that isn't given with any indexing (either from inside or outside the theory). In my day-to-day practice, I'd still think that the intersection of classes exists.
    $endgroup$
    – Pace Nielsen
    38 mins ago
















$begingroup$
I don't think it is that easy. The $n$ here is a meta-natural number (not a natural number from inside ZFC, but from my meta-theory). So there is no such formula $phi(R,n)$ of ZFC.
$endgroup$
– Pace Nielsen
42 mins ago




$begingroup$
I don't think it is that easy. The $n$ here is a meta-natural number (not a natural number from inside ZFC, but from my meta-theory). So there is no such formula $phi(R,n)$ of ZFC.
$endgroup$
– Pace Nielsen
42 mins ago












$begingroup$
That said, go ahead and replace the family I gave with a new family that isn't given with any indexing (either from inside or outside the theory). In my day-to-day practice, I'd still think that the intersection of classes exists.
$endgroup$
– Pace Nielsen
38 mins ago




$begingroup$
That said, go ahead and replace the family I gave with a new family that isn't given with any indexing (either from inside or outside the theory). In my day-to-day practice, I'd still think that the intersection of classes exists.
$endgroup$
– Pace Nielsen
38 mins ago


















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Olav Thon

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Olav Thon Fødd 29. juni 1923 Ål Yrke forretningsdrivande Olav Thon på Commons Olav Thon (fødd 29. juni 1923) er ein norsk, stort sett sjølvlært forretningsmann, fødd i Ål i Hallingdal, som har gjort store pengar på særleg kjøp, bygging og utvikling av fast eigedom, mellom anna kjøpesentre, hotell og restaurantar, i både Noreg og somme andre land. Han vart Ridder av 1. klasse av St. Olavs Orden i 2003, og er som ivrig friluftsmann æresmedlem i Den Norske Turistforening etter å ha ytt store økonomiske bidrag til foreininga sine hytter og ruter i fjellet. Han er kjend for alltid å gå med ei karakteristisk strikkelue. Olav Thon i samtale med Trond Helleland på sentralbanksjefens årstale i februar 2016 I ung alder reiste Olav Thon til Newcastle for å lære seg engelsk, og han var seinare ei tid i Sveits, der han gjorde forretningar i import og eksport. Deretter tok han til å kjøpe faste eigedomar i Noreg med stor økonomisk suksess, verksemda voks snøgt og v...
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Waikiki

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Denne artikkelen kan ha godt av ein språkvask , som reinskar opp målføringa og/eller innfører same språkstilen overalt. Denne artikkelen kan ha godt av ei opprydding for å nå ein høgare standard og/eller for å verta i tråd med standardoppsettet. Sjå korleis du redigerer ei side og stilmanualen for hjelp. Waikiki Land   USA Stat Hawaii Stad Honolulu Wikimedia Commons: Waikiki Utsyn over Waikiki Waikīkī er ein bydel i Honolulu, sør på øya Oahu, som er ei av Hawaii-øyane. Waikīkīstranda er ein av dei meste berømte strendene i verda. Waikīkī strekkjer seg langs stillehavskysten og har ei rekkje velrenommerte hotell. Bydelen er Hawaii-turismens høyborg. På Waikīkī Beach står ei statue av den lokale sportsstjerna Duka Kahanamoku. Kjelder | Denne artikkelen bygger på «Waikiki» frå Wikipedia på bokmål, den 4. november 2018. Denne artikkelen manglar kjelder eller referansar . Hjelp Wikipedia med å finna truverdige kjelder...
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Tårekanal

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Tåreapparatet ( a ) Tårekjertel ( b ) Øvre tårepunkt ( c ) Øvre tårekanalikkel ( d ) Tåresekk ( e ) Nedre tårepunkt ( f ) Nedre tårekanalikkel ( g ) Tårekanal Tårekanalen eller nase-tårekanalen (av latin ductus nasolacrimalis ) er ein ductus som frakter overflødige tårer frå tåresekken til nasehola. Opninga inn mot nedre nasegang er delvis tildekt av ei ufullstendig slimhinnefold kalla Hasners ventil. Overflødige tårer strøymer gjennom tårekanalen og trekker inn i nedre nasegang. Dette er grunnen til at ein kan få rennande nase når ein gret, og at ein somme gonger kan kjenne saltsmaken av tårer i munnen. Både tårekanalen og tåresekken er tekt med fleirlaga sylinderepitel som inneheld slimproduserande begerceller, og er begge omgjevne av bindevev. Dakryostenose (tårekanalstenose) er ei innsnevring av tårekanalen som fører til epifora, ei overflod av tårer. Medfødd dakryostenose kan føra til cystisk utviding av tårekanalen ved at det dannast ein dakryocystocele. Personar m...
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