Hall's Theorem - Proof











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I've been stuck on this proof from my textbook:



We are considering bipartite graphs only. A will refer to one of the bipartitions, and B will refer to the other.



Here are the relevant parts of the textbook



enter image description here



enter image description here



enter image description here



enter image description here



Firstly, why is $d_h(A) ge 1$ if H is a minimal subgraph that satisfies the marriage condition and contains A. If we let the degree of some vertex $a$, be greater than 1, then H would not be minimal, right? Wouldn't a subgraph H, where each vertex in A, has a degree of exactly 1, still satisfy the marriage condition and be minimal?



Also, I don't really understand the conclusion they reached. It starts out by assuming that H is a minimal subgraph that satisfies the marriage condition (and no other assumptions), and from there, it ends by saying that H does not satisfy the marriage conditions. To my understanding, they just proved



If H satisfies the marriage condition then H does not satisfy the marriage condition



So there's obviously something I'm misunderstanding here. I think, they were trying to do a proof by contradiction, but usually in those kinds of proofs, the assumption that needs to be disproved is done so by use of other premises and given information. However, our only assumption here was that H satisfies the condition. We then somehow ended up with teh conclusion that it doesn't.



Sorry for the long question.
I really appreciate the help.



Thanks






graph-theory proof-explanation bipartite-graph







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    up vote
    0
    down vote

    favorite












    I've been stuck on this proof from my textbook:



    We are considering bipartite graphs only. A will refer to one of the bipartitions, and B will refer to the other.



    Here are the relevant parts of the textbook



    enter image description here



    enter image description here



    enter image description here



    enter image description here



    Firstly, why is $d_h(A) ge 1$ if H is a minimal subgraph that satisfies the marriage condition and contains A. If we let the degree of some vertex $a$, be greater than 1, then H would not be minimal, right? Wouldn't a subgraph H, where each vertex in A, has a degree of exactly 1, still satisfy the marriage condition and be minimal?



    Also, I don't really understand the conclusion they reached. It starts out by assuming that H is a minimal subgraph that satisfies the marriage condition (and no other assumptions), and from there, it ends by saying that H does not satisfy the marriage conditions. To my understanding, they just proved



    If H satisfies the marriage condition then H does not satisfy the marriage condition



    So there's obviously something I'm misunderstanding here. I think, they were trying to do a proof by contradiction, but usually in those kinds of proofs, the assumption that needs to be disproved is done so by use of other premises and given information. However, our only assumption here was that H satisfies the condition. We then somehow ended up with teh conclusion that it doesn't.



    Sorry for the long question.
    I really appreciate the help.



    Thanks






    graph-theory proof-explanation bipartite-graph







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I've been stuck on this proof from my textbook:



      We are considering bipartite graphs only. A will refer to one of the bipartitions, and B will refer to the other.



      Here are the relevant parts of the textbook



      enter image description here



      enter image description here



      enter image description here



      enter image description here



      Firstly, why is $d_h(A) ge 1$ if H is a minimal subgraph that satisfies the marriage condition and contains A. If we let the degree of some vertex $a$, be greater than 1, then H would not be minimal, right? Wouldn't a subgraph H, where each vertex in A, has a degree of exactly 1, still satisfy the marriage condition and be minimal?



      Also, I don't really understand the conclusion they reached. It starts out by assuming that H is a minimal subgraph that satisfies the marriage condition (and no other assumptions), and from there, it ends by saying that H does not satisfy the marriage conditions. To my understanding, they just proved



      If H satisfies the marriage condition then H does not satisfy the marriage condition



      So there's obviously something I'm misunderstanding here. I think, they were trying to do a proof by contradiction, but usually in those kinds of proofs, the assumption that needs to be disproved is done so by use of other premises and given information. However, our only assumption here was that H satisfies the condition. We then somehow ended up with teh conclusion that it doesn't.



      Sorry for the long question.
      I really appreciate the help.



      Thanks






      graph-theory proof-explanation bipartite-graph







      share|cite|improve this question













      I've been stuck on this proof from my textbook:



      We are considering bipartite graphs only. A will refer to one of the bipartitions, and B will refer to the other.



      Here are the relevant parts of the textbook



      enter image description here



      enter image description here



      enter image description here



      enter image description here



      Firstly, why is $d_h(A) ge 1$ if H is a minimal subgraph that satisfies the marriage condition and contains A. If we let the degree of some vertex $a$, be greater than 1, then H would not be minimal, right? Wouldn't a subgraph H, where each vertex in A, has a degree of exactly 1, still satisfy the marriage condition and be minimal?



      Also, I don't really understand the conclusion they reached. It starts out by assuming that H is a minimal subgraph that satisfies the marriage condition (and no other assumptions), and from there, it ends by saying that H does not satisfy the marriage conditions. To my understanding, they just proved



      If H satisfies the marriage condition then H does not satisfy the marriage condition



      So there's obviously something I'm misunderstanding here. I think, they were trying to do a proof by contradiction, but usually in those kinds of proofs, the assumption that needs to be disproved is done so by use of other premises and given information. However, our only assumption here was that H satisfies the condition. We then somehow ended up with teh conclusion that it doesn't.



      Sorry for the long question.
      I really appreciate the help.



      Thanks






      graph-theory proof-explanation bipartite-graph




      graph-theory proof-explanation bipartite-graph






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      asked Dec 2 '16 at 18:54









      The_Questioner

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          The outline of the proof is




          • let $H$ be the edge minimal subgraph of $G$ that contains $A$ and satisfies the marriage condition

          • we claim it is a matching of $A$

          • suppose it is not a matching (i.e. suppose $d_H(a)>1$); then we get a contradiction


          Your paragraph "Firstly why is $d_H(a) ge 1$ ..." is basically the outline of the proof. You claim that removing one of the extra edges of $a$ does not cause the marriage condition to be violated, which is exactly what they show, except they do it by contradiction; that is, they show that if removing an extra edge of $a$ violates the marriage condition, then there is a contradiction.






          share|cite|improve this answer





















          • Thank you for the answer. Sorry for the late reply, but I was in school. another question I have is, how is it that $H - ab_1$ and $ H - ab_2$ don't satisfy the marriage conditions?
            – The_Questioner
            Dec 2 '16 at 23:48










          • @The_Questioner The point is that they do satisfy the marriage conditions. They show it by assuming those two graphs don't satisfy the marriage conditions and then arriving at a contradiction.
            – angryavian
            Dec 3 '16 at 3:13












          • But the proof says that they don't because of the minimality of H.
            – The_Questioner
            Dec 3 '16 at 3:28










          • @The_Questioner Yes. Their approach: if $H$ is minimal, then removing an edge will violate the marriage condition due to minimality of $H$; then, (some more arguments) we have a contradiction. Your approach: suppose $H$ is minimal; if $d_H(a) > 1$, then removing an extra edge won't violate the marriage condition (needs justification), which contradicts minimality of $H$.
            – angryavian
            Dec 3 '16 at 3:43











          Your Answer





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          up vote
          0
          down vote













          The outline of the proof is




          • let $H$ be the edge minimal subgraph of $G$ that contains $A$ and satisfies the marriage condition

          • we claim it is a matching of $A$

          • suppose it is not a matching (i.e. suppose $d_H(a)>1$); then we get a contradiction


          Your paragraph "Firstly why is $d_H(a) ge 1$ ..." is basically the outline of the proof. You claim that removing one of the extra edges of $a$ does not cause the marriage condition to be violated, which is exactly what they show, except they do it by contradiction; that is, they show that if removing an extra edge of $a$ violates the marriage condition, then there is a contradiction.






          share|cite|improve this answer





















          • Thank you for the answer. Sorry for the late reply, but I was in school. another question I have is, how is it that $H - ab_1$ and $ H - ab_2$ don't satisfy the marriage conditions?
            – The_Questioner
            Dec 2 '16 at 23:48










          • @The_Questioner The point is that they do satisfy the marriage conditions. They show it by assuming those two graphs don't satisfy the marriage conditions and then arriving at a contradiction.
            – angryavian
            Dec 3 '16 at 3:13












          • But the proof says that they don't because of the minimality of H.
            – The_Questioner
            Dec 3 '16 at 3:28










          • @The_Questioner Yes. Their approach: if $H$ is minimal, then removing an edge will violate the marriage condition due to minimality of $H$; then, (some more arguments) we have a contradiction. Your approach: suppose $H$ is minimal; if $d_H(a) > 1$, then removing an extra edge won't violate the marriage condition (needs justification), which contradicts minimality of $H$.
            – angryavian
            Dec 3 '16 at 3:43















          up vote
          0
          down vote













          The outline of the proof is




          • let $H$ be the edge minimal subgraph of $G$ that contains $A$ and satisfies the marriage condition

          • we claim it is a matching of $A$

          • suppose it is not a matching (i.e. suppose $d_H(a)>1$); then we get a contradiction


          Your paragraph "Firstly why is $d_H(a) ge 1$ ..." is basically the outline of the proof. You claim that removing one of the extra edges of $a$ does not cause the marriage condition to be violated, which is exactly what they show, except they do it by contradiction; that is, they show that if removing an extra edge of $a$ violates the marriage condition, then there is a contradiction.






          share|cite|improve this answer





















          • Thank you for the answer. Sorry for the late reply, but I was in school. another question I have is, how is it that $H - ab_1$ and $ H - ab_2$ don't satisfy the marriage conditions?
            – The_Questioner
            Dec 2 '16 at 23:48










          • @The_Questioner The point is that they do satisfy the marriage conditions. They show it by assuming those two graphs don't satisfy the marriage conditions and then arriving at a contradiction.
            – angryavian
            Dec 3 '16 at 3:13












          • But the proof says that they don't because of the minimality of H.
            – The_Questioner
            Dec 3 '16 at 3:28










          • @The_Questioner Yes. Their approach: if $H$ is minimal, then removing an edge will violate the marriage condition due to minimality of $H$; then, (some more arguments) we have a contradiction. Your approach: suppose $H$ is minimal; if $d_H(a) > 1$, then removing an extra edge won't violate the marriage condition (needs justification), which contradicts minimality of $H$.
            – angryavian
            Dec 3 '16 at 3:43













          up vote
          0
          down vote










          up vote
          0
          down vote









          The outline of the proof is




          • let $H$ be the edge minimal subgraph of $G$ that contains $A$ and satisfies the marriage condition

          • we claim it is a matching of $A$

          • suppose it is not a matching (i.e. suppose $d_H(a)>1$); then we get a contradiction


          Your paragraph "Firstly why is $d_H(a) ge 1$ ..." is basically the outline of the proof. You claim that removing one of the extra edges of $a$ does not cause the marriage condition to be violated, which is exactly what they show, except they do it by contradiction; that is, they show that if removing an extra edge of $a$ violates the marriage condition, then there is a contradiction.






          share|cite|improve this answer












          The outline of the proof is




          • let $H$ be the edge minimal subgraph of $G$ that contains $A$ and satisfies the marriage condition

          • we claim it is a matching of $A$

          • suppose it is not a matching (i.e. suppose $d_H(a)>1$); then we get a contradiction


          Your paragraph "Firstly why is $d_H(a) ge 1$ ..." is basically the outline of the proof. You claim that removing one of the extra edges of $a$ does not cause the marriage condition to be violated, which is exactly what they show, except they do it by contradiction; that is, they show that if removing an extra edge of $a$ violates the marriage condition, then there is a contradiction.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '16 at 19:34









          angryavian

          37.5k13179




          37.5k13179












          • Thank you for the answer. Sorry for the late reply, but I was in school. another question I have is, how is it that $H - ab_1$ and $ H - ab_2$ don't satisfy the marriage conditions?
            – The_Questioner
            Dec 2 '16 at 23:48










          • @The_Questioner The point is that they do satisfy the marriage conditions. They show it by assuming those two graphs don't satisfy the marriage conditions and then arriving at a contradiction.
            – angryavian
            Dec 3 '16 at 3:13












          • But the proof says that they don't because of the minimality of H.
            – The_Questioner
            Dec 3 '16 at 3:28










          • @The_Questioner Yes. Their approach: if $H$ is minimal, then removing an edge will violate the marriage condition due to minimality of $H$; then, (some more arguments) we have a contradiction. Your approach: suppose $H$ is minimal; if $d_H(a) > 1$, then removing an extra edge won't violate the marriage condition (needs justification), which contradicts minimality of $H$.
            – angryavian
            Dec 3 '16 at 3:43


















          • Thank you for the answer. Sorry for the late reply, but I was in school. another question I have is, how is it that $H - ab_1$ and $ H - ab_2$ don't satisfy the marriage conditions?
            – The_Questioner
            Dec 2 '16 at 23:48










          • @The_Questioner The point is that they do satisfy the marriage conditions. They show it by assuming those two graphs don't satisfy the marriage conditions and then arriving at a contradiction.
            – angryavian
            Dec 3 '16 at 3:13












          • But the proof says that they don't because of the minimality of H.
            – The_Questioner
            Dec 3 '16 at 3:28










          • @The_Questioner Yes. Their approach: if $H$ is minimal, then removing an edge will violate the marriage condition due to minimality of $H$; then, (some more arguments) we have a contradiction. Your approach: suppose $H$ is minimal; if $d_H(a) > 1$, then removing an extra edge won't violate the marriage condition (needs justification), which contradicts minimality of $H$.
            – angryavian
            Dec 3 '16 at 3:43
















          Thank you for the answer. Sorry for the late reply, but I was in school. another question I have is, how is it that $H - ab_1$ and $ H - ab_2$ don't satisfy the marriage conditions?
          – The_Questioner
          Dec 2 '16 at 23:48




          Thank you for the answer. Sorry for the late reply, but I was in school. another question I have is, how is it that $H - ab_1$ and $ H - ab_2$ don't satisfy the marriage conditions?
          – The_Questioner
          Dec 2 '16 at 23:48












          @The_Questioner The point is that they do satisfy the marriage conditions. They show it by assuming those two graphs don't satisfy the marriage conditions and then arriving at a contradiction.
          – angryavian
          Dec 3 '16 at 3:13






          @The_Questioner The point is that they do satisfy the marriage conditions. They show it by assuming those two graphs don't satisfy the marriage conditions and then arriving at a contradiction.
          – angryavian
          Dec 3 '16 at 3:13














          But the proof says that they don't because of the minimality of H.
          – The_Questioner
          Dec 3 '16 at 3:28




          But the proof says that they don't because of the minimality of H.
          – The_Questioner
          Dec 3 '16 at 3:28












          @The_Questioner Yes. Their approach: if $H$ is minimal, then removing an edge will violate the marriage condition due to minimality of $H$; then, (some more arguments) we have a contradiction. Your approach: suppose $H$ is minimal; if $d_H(a) > 1$, then removing an extra edge won't violate the marriage condition (needs justification), which contradicts minimality of $H$.
          – angryavian
          Dec 3 '16 at 3:43




          @The_Questioner Yes. Their approach: if $H$ is minimal, then removing an edge will violate the marriage condition due to minimality of $H$; then, (some more arguments) we have a contradiction. Your approach: suppose $H$ is minimal; if $d_H(a) > 1$, then removing an extra edge won't violate the marriage condition (needs justification), which contradicts minimality of $H$.
          – angryavian
          Dec 3 '16 at 3:43


















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