 variables; together, they take values in the product space $2^{\mathbb{N}}$.  Suppose we have a collection of statements $S_t$ in these boolean variables such that every finite subset is satisfiable. Then I claim that we can prove that they are all simultaneously satisfiable by using a compactness argument.
Let $F$ be a finite set. Then the set of all truth assignments (this is a subset of $2^{\mathbb{N}}$) which satisfy $S_t$ for $t \in F$ is a closed set $V_F$ of assignments satisfying the sentences in $F$.  The intersection of any finitely many of the $V_F$ is nonempty, so by the finite intersection property, the intersection of all of them is nonempty (since the product space is compact), whence any truth in this intersection satisfies all the statements.
I don't know how this works in predicate logic.

A:

Lemma: A topological space $X$ is compact if and only if for every collection $\mathcal{C}$ of closed sets with the finite intersection property has nonempty intersection over the collection.
Proposition: $\mathbb{M}(\mathcal{L})$ is compact if and only if every finitely satisfiable $\mathcal{L}$-theory is satisfiable.
Proof: Consider the space $\mathbb{M}(\mathcal{L})=\{\Phi \; | \; \Phi \; \text{is a maximal} \; \mathcal{L} \text{-theory}\}$. For each $\mathcal{L}$-sentence $\varphi$ let $[(\varphi)]=\{ \Phi \in \mathbb{M}(\mathcal{L}) \; | \; \varphi \in \Phi\}$. A subbase $\cal{B}$ for a topology on $\mathbb{M}(\mathcal{L})$ is given by the sets $[(\varphi)]$. That is, the open subsets of $\mathbb{M}(\mathcal{L})$ are the union of the finite intersections of elements of $\mathcal{B}$. To see that $\mathcal{B}$ defines a topology on note that $[(\forall x(x\neq x))]=\emptyset$ and $[(\forall x(x=x))]=\mathbb{M}(\mathcal{L})$. Furthermore, arbitrary unions are already defined to be in the topology and finite intersections are in the topology since $[(\varphi)]\cap [(\theta)]=[(\varphi \wedge \theta)]$, which is defined to be open. 
Assume logical compact
variables; together, they take values in the merchandise quad $ 2^{\mathbb{N}}$.   Suppose we have a collection of argument $ S_t$ in these boolean variables such that every finite subset is satisfiable. Then I claim that we can raise that they are wholly simultaneously satisfiable by using a compactness argument. 
 Let $ F$ be a finite set. Then the hardening of all truth assignments (this is a subset of $ 2^{\mathbb{N}}$) which meet $ S_t$ for $ t \in F$ is a closed hardening $ V_F$ of assignments satisfying the sentence in $ F$.   The intersection of any finitely many of the $ V_F$ is nonempty, so by the finite intersection property, the overlap of all of them is nonempty (since the product space is compendious), whence any truth in this overlap satisfies all the statements. 
 I don't know how this works in predicate logic. 

 A: 

 Lemma: A topological space $ X$ is compact if and only if for every solicitation $ \mathcal{C}$ of closed sets with the finite intersection property has nonempty intersection over the solicitation. 
 Proposition: $ \mathbb{M}(\mathcal{L})$ is compendious if and entirely if every finitely satisfiable $ \mathcal{L}$-theory is satisfiable. 
 Proof: Consider the space $ \mathbb{M}(\mathcal{L})=\{\Phi \; | \; \Phi \; \text{is a maximal } \; \mathcal{L } \text{-theory}\}$. For each $ \mathcal{L}$-sentence $ \varphi$ let $ [ (\varphi)]=\ { \Phi \in \mathbb{M}(\mathcal{L }) \; | \; \varphi \in \Phi\}$. A subbase $ \cal{B}$ for a topology on $ \mathbb{M}(\mathcal{L})$ is given by the sets $ [ (\varphi)]$. That is, the exposed subsets of $ \mathbb{M}(\mathcal{L})$ are the union of the finite intersection of chemical element of $ \mathcal{B}$. To see that $ \mathcal{B}$ defines a topology on note that $ [ (\forall x(x\neq x))]=\emptyset$ and $ [ (\forall x(x = x))]=\mathbb{M}(\mathcal{L})$. Furthermore, arbitrary unions are already defined to be in the topology and finite intersection are in the topology since $ [ (\varphi)]\cap [ (\theta)]=[(\varphi \wedge \theta)]$, which is defined to be open. 
 Assume logical covenant
 vagiables; together, they taye values in thg producv space $2^{\mathbb{N}}$.  Suppose we have a collectiln of suctements $S_t$ in these boolean nariables suci that every finmfe subscc is awtisyieble. Then I clalm that we wan prove that tfev are all simultaneously satisfiable by usimg a compactness arglmqnt.
Lsn $N$ be a finite set. Then the set or all tguth assignments (yhis is a subset of $2^{\mathbb{J}}$) whlch satisfy $S_t$ for $t \in F$ is q cljwed set $V_F$ ow assignments satisfyihg the sentences in $F$.  The intefsectnon of any dibitfny many of vhe $V_F$ is nonempty, so by tve finiye intersectiok pro'ertt, the intersection of all of them is nogempty (sitcz the product space iw xompawt), wvencd anh tduvh jn thid iitersection satisfies qll the statements.
I djb't know how tgis wowkf in predicate logic.

A:

Lemma: A topologican slace $X$ is compact if ane only if for every clllection $\mathcal{C}$ of closed sets with the finite intersecdion 'rupexbn har nlnempty intersection over the collection.
Propofjtook: $\mathbb{M}(\mathcal{J})$ is compacy lf wnd only if exery fnhifely satisfiable $\mwthcal{L}$-jheory is satissiabke.
Proof: Consider the space $\nathbb{M}(\mathccl{L})=\{\Phi \; | \; \Phi \; \text{is a maximal} \; \matncal{L} \text{-theory}\}$. For each $\machcal{L}$-aentence $\vagphi$ let $[(\bxrphi)]=\{ \Phi \in \matfbb{K}(\mdthcal{L}) \; | \; \varphi \in \Phi\}$. A subbase $\ral{B}$ yor a toooloby on $\iathbb{M}(\matjcal{L})$ is given by the sehs $[(\vatphi)]$. Tvat is, the open subsets of $\mathbb{M}(\mathcal{L})$ are the uniom mf nhe finitz intevsections of elqments of $\matheal{B}$. To fee tfat $\mathcam{B}$ defiies a topolody on note thdj $[(\forall x(x\nex x))]=\emptyfet$ qnd $[(\dorall b(b=x))]=\mathbb{M}(\mathcsl{L})$. Furthvriore, arbitrary unions ere xmready defined cu be in the topokogh agd fmnite hntersectionv ard iv the gopology sikce $[(\varlhi)]\cap [(\theta)]=[(\varphi \wadge \theta)]$, which is degiked to be open. 
Asfume logical vompact
variables; together, they take values in the $2^{\mathbb{N}}$. we have collection of statements such every finite subset satisfiable. Then I that we can prove that they all simultaneously satisfiable by using a compactness argument. Let $F$ be a finite Then the set of all truth assignments (this is a subset of $2^{\mathbb{N}}$) satisfy for \in is a closed set $V_F$ of assignments satisfying the sentences in $F$. The intersection of any many of the $V_F$ is nonempty, so by finite intersection property, the of all of them is (since product space compact), any in this intersection all the statements. I don't know how this works in predicate logic. A: Lemma: A topological space is compact only if every $\mathcal{C}$ closed sets with intersection property has nonempty intersection over $\mathbb{M}(\mathcal{L})$ is compact if and only if every satisfiable $\mathcal{L}$-theory satisfiable. Proof: Consider the space $\mathbb{M}(\mathcal{L})=\{\Phi | \; \Phi \; \text{is a maximal} \; \text{-theory}\}$. For each $\mathcal{L}$-sentence $\varphi$ let $[(\varphi)]=\{ \Phi \in \mathbb{M}(\mathcal{L}) \; | \; \varphi \in subbase $\cal{B}$ for a on $\mathbb{M}(\mathcal{L})$ is by sets That the open of $\mathbb{M}(\mathcal{L})$ are the union of the finite intersections of elements $\mathcal{B}$. To see that $\mathcal{B}$ defines a topology on note x(x\neq and $[(\forall x(x=x))]=\mathbb{M}(\mathcal{L})$. arbitrary unions are already to in the topology and are the [(\theta)]=[(\varphi \theta)]$, is defined to be Assume logical compact
 variables; together, they take Values in thE prodUct SpaCe $2^{\MathBb{N}}$.  SUppose we have a cOLlecTion of statements $S_t$ in thEse boOlEAn vaRIaBles sUch that EVeRY FinItE sUbsEt IS sAtisfIabLe. Then I Claim that wE caN pRove that they ARe All simultaNeoUsly satisfiaBle By usinG a ComPActneSs aRgumeNt.
Let $F$ BE a finiTe set. Then ThE Set of aLL truth aSSIgNmenTs (this is a subset of $2^{\MAtHBb{N}}$) which satisfY $S_t$ for $T \iN f$ iS A CloSed Set $V_F$ of assIgNmentS SatisfyINg THE SenTEnces in $F$.  The inTersection oF Any FiniteLy ManY Of the $V_f$ is noNeMPty, So by the finiTe inTersectioN propeRTy, the inTErsectiOn of alL of TheM is nONeMpTy (sInCE thE PrOduCT spAce is comPaCt), WhencE any TRUTH in tHis InteRsectIon satisfies aLl tHe stATemEnts.
I Don't kNow hOw This wOrks in PrediCaTe logic.

A:

Lemma: A tOpolOgical spaCe $X$ Is ComPaCt if aND only iF foR evEry collEction $\mAThcAl{c}$ OF ClOsed sets with the finItE INtErsectioN propeRTy HaS Nonempty InTerSectION over The cOLlEction.
PrOpositIOn: $\MaThbb{M}(\maThCal{L})$ is CoMpaCt iF and oNLy if Every fInitely sAtisfIAble $\mathcal{L}$-thEOry is satisfiaBLe.
pROoF: consIdeR the space $\maThbb{m}(\MathCal{L})=\{\pHi \; | \; \phi \; \TExt{is A maxiMaL} \; \MaTHcal{L} \text{-theory}\}$. For eAcH $\mathcAl{L}$-seNtence $\varphi$ lEt $[(\varphi)]=\{ \PhI \IN \Mathbb{M}(\mAthcAL{L}) \; | \; \VArphi \in \Phi\}$. A subBase $\cAl{B}$ for a topOLogy on $\maThbb{M}(\Mathcal{L})$ Is given by THE sets $[(\varPhi)]$. thaT is, The OPEn Subsets of $\mathBB{m}(\matHcAl{L})$ are tHe uNion of tHe fIniTe iNteRsEctions of Elements Of $\MaThCaL{B}$. TO see tHAt $\mathcaL{B}$ DefInEs a TopolOGy on noTe thaT $[(\forAlL x(X\Neq X))]=\emptysET$ aND $[(\ForaLl X(x=X))]=\matHbb{m}(\mAthcaL{L})$. FuRTheRmore, arBitrary unIonS Are aLrEaDy definEd to be in the toPoLogy and finItE inTersecTIOns are in The topology since $[(\varphi)]\cAP [(\theta)]=[(\vArpHi \wedGe \thEta)]$, which iS deFined tO be OPen. 
AssUme logIcal cOmPacT
 variables; together, they  take valu es in  th e p ro duct  spa ce $2^{\mathbb { N}}$ .  Suppose we have a c ollec ti o n of st ateme nts $S_ t $i n  th es eboo le a nvaria ble s suchthat every  fi ni te subset is sa tisfiable.  Th en I claim t hat  we ca npro v e tha t t hey a re all simult aneouslysa t isfiab l e by us i n ga co mpactness argumen t .L et $F$ be a fi nite s et .  T h e n t heset of all  t rutha ssignme n ts ( t his is a subset o f $2^{\math b b{N }}$) w hi chs atisfy  $S_t $f or$t \in F$ i s aclosed se t $V_F $  of ass i gnments  satis fyi ngthes en te nce si n $ F $.   T h e i ntersect io nof an y fi n i t e ly m any  ofthe $ V_F$ is nonem pty , so bythe f inite  int er secti on pro perty ,the intersectio n of  all of t hem  i s n on empty (since  th e p roductspace i s  co mp a c t ),  whence any truthin t hi s inters ection sa ti s fies all  t hestat e m ents. 
I d o n' t know h ow thi s  w or ks in p re dicate  l ogi c.
A:
L emma : A to pologica l spa c e $X$ is compa c t if and only if f or ever y c ollection $ \mat h cal{ C}$o fclo s ed se ts wi th th e  finite intersectio nproper ty ha s nonempty in tersection o v er the c olle c ti o n.
Proposition : $\m athbb{M}(\ m athcal{L })$ i s compac t if ando n ly if ev ery  fi nit ely s at isfiable $\ma t h cal{ L} $-theor y i s satis fia ble .
P roo f:  Consider  the spa ce  $ \m at hbb {M}(\ m athcal{L }) =\{ \P hi\; |\ ; \Phi  \; \ text {i sa  ma ximal}\ ;\ m athc al {L } \t ext {- theor y}\} $ . F or each  $\mathca l{L } $-se nt en ce $\va rphi$ let $[( \v arphi)]=\{  \ Phi  \in \ m a thbb{M}( \mathcal{L}) \; | \; \v a rphi \i n \ Phi\} $. A  subbase$\c al{B}$  fo r  a top ologyon $\ ma thb b { M}(\m a t hc al{ L} )$ is give n bythe s et s $[ (\varph i)]$. That is, the ope n subsets of$\m athb b { M} (\m a th c al{ L} ) $ a r e  the union of t he finitein t er sections o f  el em ents of  $\math cal{B } $. To s ee that $ \mathcal{ B} $ de f i nes  a topolog y on not e that $[ ( \fora l lx(x\n eqx))]=\ em pty set$and $[ ( \fo rallx(x=x) )] =\math bb{M} (\ mathcal{ L})$. Furthermore, arbi traryunion s a re alread y d e fin ed to bein t he topolog y a ndfinit e i n terse ctio n sare in th e to p ology sin c e$[( \ v ar phi)]\cap [ ( \ t het a)]=[ (\v a rphi \ wedg e \theta)]$, whic h  is defined to  beo p en.  
A s sume  l ogical compact 
 variables;_together, they_take values in the_product space_$2^{\mathbb{N}}$._ Suppose_we_have a collection_of statements $S_t$_in these boolean variables_such that every_finite_subset is satisfiable. Then I claim that we can prove that they are all_simultaneously_satisfiable by_using_a_compactness argument.
Let $F$ be a_finite set. Then the set_of all_truth assignments (this is a subset of $2^{\mathbb{N}}$)_which_satisfy $S_t$ for_$t \in F$ is a closed set $V_F$ of_assignments satisfying the sentences in $F$._ The intersection_of_any_finitely many of the_$V_F$ is nonempty, so by the_finite intersection property, the intersection of_all of them is nonempty (since the_product space is compact), whence any_truth in this intersection satisfies_all the_statements.
I don't know how this_works in predicate_logic.

A:

Lemma: A_topological space $X$_is compact if and only if_for every collection_$\mathcal{C}$ of closed sets with the_finite_intersection property has_nonempty_intersection_over the_collection.
Proposition: $\mathbb{M}(\mathcal{L})$ is_compact_if and_only_if every finitely satisfiable $\mathcal{L}$-theory is_satisfiable.
Proof:_Consider the space $\mathbb{M}(\mathcal{L})=\{\Phi \; | \;_\Phi \; \text{is a_maximal}_\; \mathcal{L} \text{-theory}\}$. For_each $\mathcal{L}$-sentence $\varphi$ let $[(\varphi)]=\{_\Phi \in \mathbb{M}(\mathcal{L}) \; | \;_\varphi \in_\Phi\}$. A_subbase $\cal{B}$ for a topology on $\mathbb{M}(\mathcal{L})$ is given by the_sets $[(\varphi)]$. That is, the open_subsets of $\mathbb{M}(\mathcal{L})$ are_the union_of_the finite intersections_of_elements of_$\mathcal{B}$. To see that $\mathcal{B}$ defines a_topology on_note that $[(\forall x(x\neq x))]=\emptyset$ and_$[(\forall x(x=x))]=\mathbb{M}(\mathcal{L})$. Furthermore, arbitrary_unions_are already defined to be in_the topology and finite intersections are_in the topology since $[(\varphi)]\cap_[(\theta)]=[(\varphi_\wedge_\theta)]$, which is defined to_be open. 
Assume logical compact