 also infinite is not shown. Sure our $\Gamma$ already has an infinite model, but nothing says that its enumerable model must be identical to its infinite model.
I can't see how this claim is true, but I guess it's supposed to be true?

A:

The (downward) [Löwenheim-Skolem] theorem actually does say that if $\Gamma$ has an infinite model, there also is a countably infinite model (so of the cardinality of $\Bbb N$).

A:

As the other answers have said, there are various formulations of the downward Lowenheim-Skolem theorem which are ultimately equivalent. I'll assume - like Alex Kruckman - that you mean the version "If $T$ is a satisfiable theory of size $\le\kappa$ with $\kappa$ infinite, then $T$ has a model of size $\le\kappa$."

Answer
Suppose $T$ is a countable theory and $M$ is an infinite model of $T$. Consider the larger theory $$T'=T\cup\{\exists x_1,...,x_n(\bigwedge_{1\le i<j\le n}x_i\not=x_j):n\in\mathbb{N}\}.$$ The additional sentences in $T'$ say "the domain has size at least $n$."
Since $M\models T$ and $M$ is infinite, we have $M\models T'$. Now apply downward Lowenheim-Skolem to the theory $T'$. We get a finite-or-countable $M'\models T'$. But clearly every model of $T'$ is infinite, so $M'$ is countably infinite.

Actually, we can do even better in a sense: rather than stopping at $T'$, we can "go all the way" and move from $T'$ to the full elementary diagram (or theory) of $M$: $$Th(M)=\{\varphi:M\models\varphi\}.$$ While $Th(M)$ is in one sense overkill (there are possibly many sentences in $Th(M)$ which aren't in $T'$), in another sense it's ultimately better since the assignment $M\mapsto Th(M)$ is an absolutely crucial thing in model theory.

The "right" Lowenheim-Skolem theorem
Where the "expand-the-theory" trick above breaks down is when
also infinite is not shown. Sure our $ \Gamma$ already has an countless exemplar, but nothing says that its enumerable exemplar must be identical to its countless model. 
 I can't see how this claim is truthful, but I guess it's supposed to be truthful? 

 A: 

 The (downward) [ Löwenheim - Skolem ] theorem actually does say that if $ \Gamma$ suffer an infinite mannequin, there besides is a countably infinite model (so of the cardinality of $ \Bbb N$). 

 deoxyadenosine monophosphate: 

 As the other answers have say, there are various formulations of the down Lowenheim - Skolem theorem which are ultimately equivalent. I'll assume - like Alex Kruckman - that you mean the adaptation " If $ T$ is a satisfiable hypothesis of size $ \le\kappa$ with $ \kappa$ infinite, then $ T$ own a model of size $ \le\kappa$. " 

 Answer 
 Suppose $ T$ is a countable theory and $ M$ is an countless model of $ T$. Consider the larger theory $ $ T'=T\cup\{\exists x_1,... ,x_n(\bigwedge_{1\le i < j\le n}x_i\not = x_j):n\in\mathbb{N}\}.$$ The additional sentences in $ T'$ read " the sphere has size at least $ n$. " 
 Since $ M\models T$ and $ M$ is infinite, we have $ M\models T'$. Now apply downward Lowenheim - Skolem to the hypothesis $ T'$. We get a finite - or - countable $ M'\models T'$. But intelligibly every model of $ T'$ is infinite, so $ M'$ is countably countless. 

 Actually, we can dress even better in a sense: preferably than stopping at $ T'$, we can " go all the way " and move from $ T'$ to the wide elementary diagram (or theory) of $ M$: $ $ Th(M)=\{\varphi: M\models\varphi\}.$$ While $ Th(M)$ is in one sense overkill (there are possibly many sentences in $ Th(M)$ which aren't in $ T'$), in another sense it's ultimately better since the appointment $ M\mapsto Th(M)$ is an absolutely crucial thing in model theory. 

 The " correct " Lowenheim - Skolem theorem 
 Where the " boom - the - theory " trick above breaks down is when
 aldo infinite is not shown. Sure our $\Gamma$ alreadb has ah infinige model, but nothing says thet irs enymerable model must be identicap to its infmnite model.
I can'v see how this glaim ms true, but I goess it's sup[osed to be trge?

X:

Tke (downward) [Löwenheim-Skolem] theorem astually dles say that is $\Gakia$ hzs an infinite model, there also is a counuably infinite modrl (so of the cardinality ov $\Bbh N$).

A:

As the other ajswers have saiq, there are vxrious formulations of the downward Lowenheim-Skolem tfeorek which arg hlhhmately equmvalenn. I'll assume - like Alax Kruclman - that you meen tye version "If $T$ is a vatisfiable theory of size $\ne\iappa$ with $\kappa$ undinitg, thet $T$ fqs x mkdxl kf sizf $\lx\kappa$."

Answed
Suppose $T$ us a countable theoty qnd $M$ is an ihfinitq iodel of $T$. Consider the larger theory $$T'=U\cup\{\esists x_1,...,x_n(\bigwedge_{1\le i<j\lw n}x_i\not=x_j):n\in\mathbb{N}\}.$$ Jhe additijnal sentences in $T'$ say "the domain has size at ledst $n$."
Aknct $N\modeuw H$ and $M$ is infinite, we have $M\models T'$. Now appjg coenward Lowenhelm-Skolem to the throgy $J'$. We get a finkte-or-counfable $M'\models T'$. Buh clearjy evwry model of $Y'$ is infinite, so $M'$ is countqbly infinitv.

Actyally, we can do evzn better in a sgnse: rsther than stopping at $C'$, we czn "go all tje way" ans move from $T'$ to ghe fgll elemtvtary diagram (or eheory) of $M$: $$Tk(M)=\{\varphi:O\modgls\varpri\}.$$ While $Tj(M)$ is in one sense overklll (tkere dre possibpy many sentences in $Th(M)$ which ecen't in $T'$), in snmthvr sense nt's ulbimately better since the assngnment $I\mapsgo Th(M)$ is zn absonutely crucyal thing in kldel theory.

Tie "right" Jowebhein-Skolem gheorem
Where tne "expand-nht-theory" truck above breaks dpwn js when
also infinite is not shown. Sure our has infinite model, nothing says that identical its infinite model. can't see how claim is true, but I guess supposed to be true? A: The (downward) [Löwenheim-Skolem] theorem actually does say that $\Gamma$ has an infinite model, there also is a countably infinite model (so the of N$). As the other answers have said, there are various formulations of the downward Lowenheim-Skolem theorem which ultimately equivalent. I'll assume - like Alex Kruckman that you mean the "If $T$ is a satisfiable of $\le\kappa$ with infinite, $T$ a model of $\le\kappa$." Answer Suppose $T$ is a countable theory and $M$ is an infinite model of $T$. Consider larger theory i<j\le n}x_i\not=x_j):n\in\mathbb{N}\}.$$ additional in say "the domain at least $n$." Since $M\models T$ infinite, we have $M\models T'$. Now apply downward to the $T'$. We get a finite-or-countable $M'\models But clearly every model of $T'$ is infinite, $M'$ is countably infinite. Actually, we can do even better in a sense: rather than $T'$, we can "go the way" and from to full diagram (or of $M$: $$Th(M)=\{\varphi:M\models\varphi\}.$$ While $Th(M)$ is in one sense overkill (there possibly many sentences in $Th(M)$ which aren't in $T'$), in it's better since the $M\mapsto Th(M)$ is an crucial in model theory. The theorem the breaks is
 also infinite is not shown. SurE our $\Gamma$ aLreadY haS an InFiniTe moDel, but nothing sAYs thAt its enumerable model muSt be iDeNTicaL To Its inFinite mODeL.
i Can'T sEe How ThIS cLaim iS trUe, but I gUess it's supPosEd To be true?

A:

The (DOwNward) [LöwenHeiM-Skolem] theorEm aCtuallY dOes SAy thaT if $\gamma$ Has an iNFinite Model, therE aLSo is a cOUntably INFiNite Model (so of the cardiNAlITy of $\Bbb N$).

A:

As the Other aNsWErS HAve SaiD, there are vArIous fORmulatiONs OF THe dOWnward LowenheIm-Skolem theORem Which aRe UltIMately EquivAlENt. I'Ll assume - likE AleX Kruckman - That yoU Mean the VErsion "IF $T$ is a sAtiSfiAble THeOrY of SiZE $\le\KApPa$ wITh $\kAppa$ infiNiTe, Then $T$ Has a MODEL of sIze $\Le\kaPpa$."

AnSwer
Suppose $T$ iS a cOuntABle TheorY and $M$ Is an InFinitE model Of $T$. CoNsIder the larger thEory $$t'=T\cup\{\exisTs x_1,...,X_n(\BigWeDge_{1\le I<J\le n}x_i\Not=X_j):n\In\mathbB{N}\}.$$ The adDItiOnAL SEnTences in $T'$ say "the domAiN HAs Size at leAst $n$."
SiNCe $m\mODels T$ and $m$ iS inFiniTE, We havE $M\moDElS T'$. Now appLy downWArD LOwenheiM-SKolem tO tHe tHeoRy $T'$. We GEt a fInite-oR-countabLe $M'\moDEls T'$. But clearly EVery model of $T'$ iS InFINiTE, so $M'$ Is cOuntably infInitE.

actuAlly, WE cAn dO Even bEtter In A SeNSe: rather than stoppinG aT $T'$, we caN "go alL the way" and movE from $T'$ to thE FULl elemenTary DIaGRam (or theory) of $M$: $$th(M)=\{\vaRphi:M\modelS\Varphi\}.$$ WhIle $Th(m)$ is in one Sense overKILl (there aRe pOssIblY maNY SeNtences in $Th(M)$ wHICh arEn'T in $T'$), in aNotHer sensE it'S ulTimAteLy Better sinCe the assIgNmEnT $M\MapSto Th(m)$ Is an absoLuTelY cRucIal thINg in moDel thEory.

thE "rIGht" lowenheIM-SKOLem tHeOrEm
WhEre ThE "expaNd-thE-TheOry" tricK above breAks DOwn iS wHeN
 also infinite is not show n. Sure ou r $\G amm a$al read y ha s an infinitem odel , but nothing says tha t its  e n umer a bl e mod el must be i den ti ca l t oi ts  infi nit e model .
I can'tsee  h ow this clai m  i s true, bu t I  guess it'ssup posedto  be true? 

A :

Th e (dow n ward)[Löwenhei m- S kolem] theorem a ct uall y does say that i f  $ \ Gamma$ has aninfini te mo d e l,the re also is  a  coun t ably in f in i t e  mo d el (so of the  cardinalit y  of  $\Bbb  N $). 
A:

A s the  o t her  answers ha ve s aid, ther e arev ariousf ormulat ions o f t hedown w ar dLow en h eim - Sk ole m  th eorem wh ic hare u ltim a t e l y eq uiv alen t. I' ll assume - l ike  Ale x  Kr uckma n - t hatyo u mea n theversi on  "If $T$ is a s atis fiable th eor yofsi ze $\ l e\kapp a$wit h $\kap pa$ inf i nit e, t h en  $T$ has a model o fs i ze  $\le\ka ppa$." 
A ns w er
Suppo se  $T $ is a  coun tabl e  t heory an d $M$i san  infini te  model  o f $ T$.  Cons i derthe la rger the ory $ $ T'=T\cup\{\exi s ts x_1,...,x_ n (\ b i gw e dge_ {1\ le i<j\le n }x_i \ not= x_j) : n\ in\ m athbb {N}\} .$ $  T h e additional senten ce s in $ T'$ s ay "the domai n has size a t  least $ n$." Si n ce $M\models T $ and  $M$ is in f inite, w e hav e $M\mod els T'$.N o w applydow nwa rdLow e n he im-Skolem tot h e th eo ry $T'$ . W e get a  fi nit e-o r-c ou ntable $M '\models  T '$ .Bu t c learl y  every m od elof  $T '$ is infini te, s o $M '$  i s  co untably in f i nite .
A ctua lly ,we ca n do eve n bette r in a se nse :  rat he rthan st opping at $T' $,  we can "g oall  the w a y " and mo ve from $T'$ to the ful l  elemen tar y dia gram  (or theo ry)  of $M $:$ $Th(M) =\{\va rphi: M\ mod e l s\var p h i\ }.$ $While $Th( M ) $ i s inon e se nse ove rkill (there are p o ssi bly many sent enc es i n $T h(M ) $w hic ha ren ' t  in $T'$), in a nother sen se it 's ultimat e lybe tter si nce the  assi g nment $ M\mapstoTh(M)$ is  a n ab s o lut ely crucia l thingin modelt heory . 
The " rig ht" Lo we nhe im-Sk olem t h eor em
Wh ere th e"expan d-the -t heory" t rick above breaks downis whe n
 also_infinite is_not shown. Sure our_$\Gamma$ already_has_an infinite_model,_but nothing says_that its enumerable_model must be identical_to its infinite_model.
I_can't see how this claim is true, but I guess it's supposed to be_true?

A:

The_(downward) [Löwenheim-Skolem]_theorem_actually_does say that if $\Gamma$_has an infinite model, there_also is_a countably infinite model (so of the cardinality_of_$\Bbb N$).

A:

As the_other answers have said, there are various formulations of_the downward Lowenheim-Skolem theorem which are_ultimately equivalent. I'll_assume_-_like Alex Kruckman -_that you mean the version "If_$T$ is a satisfiable theory of_size $\le\kappa$ with $\kappa$ infinite, then $T$_has a model of size $\le\kappa$."

Answer
Suppose_$T$ is a countable theory_and $M$_is an infinite model of_$T$. Consider the_larger theory_$$T'=T\cup\{\exists x_1,...,x_n(\bigwedge_{1\le i<j\le_n}x_i\not=x_j):n\in\mathbb{N}\}.$$ The additional sentences in $T'$_say "the domain_has size at least $n$."
Since $M\models_T$_and $M$ is_infinite,_we_have $M\models_T'$. Now apply_downward_Lowenheim-Skolem to_the_theory $T'$. We get a finite-or-countable_$M'\models_T'$. But clearly every model of $T'$_is infinite, so $M'$_is_countably infinite.

Actually, we can_do even better in a_sense: rather than stopping at $T'$,_we can_"go all_the way" and move from $T'$ to the full elementary diagram_(or theory) of $M$: $$Th(M)=\{\varphi:M\models\varphi\}.$$ While_$Th(M)$ is in one_sense overkill_(there_are possibly many_sentences_in $Th(M)$_which aren't in $T'$), in another sense_it's ultimately_better since the assignment $M\mapsto Th(M)$_is an absolutely crucial_thing_in model theory.

The "right" Lowenheim-Skolem theorem
Where_the "expand-the-theory" trick above breaks down_is when