$mathcal{f}$(A $cap$ B) = $mathcal{f}$(A) $cap$ $mathcal{f}$(B) $iff$ $mathcal{f}$ is injective. Numerical...












0












$begingroup$


Need some help to check whether my understanding of a subject is right or not.



$1$. Example of injective function.



$mathcal{f}$: $mathbb{N}$ $longrightarrow$ $mathbb{N}$



$x$ $longmapsto$ $2x$



A = {$1,2,3,4$}, B = {$2,4,5,7$}, (A $cap$ B) = {$2,4$}



$mathcal{f}$(A) = {$2,4,6,8$}, $space$ $mathcal{f}$(B) = {$4,8,10,14$}



$mathcal{f}$(A $cap$ B) = $mathcal{f}$(A) $cap$ $mathcal{f}$(B) = {$4,8$} $Longrightarrow$ $mathcal{f}$ is injective



$2$. Example of a function that is not injective



$mathcal{f}$: $mathbb{R}$ $longrightarrow$ $mathbb{R}$



$x$ $longmapsto$ $x^2$



A = {$2,4,6,8$}, B = {$-4,-8,-12,-16$}, (A $cap$ B) = $lbrace$ $emptyset$ $rbrace$



$mathcal{f}$(A) = {$4,16,36,64$}, $space$ $mathcal{f}$(B) = {$16,64,144,256$}



$mathcal{f}$(A) $cap$ $mathcal{f}$(B) = {$16,64$} $neq$ $mathcal{f}$(A $cap$ B) = $lbrace$ $emptyset$ $rbrace$ $Longrightarrow$ $mathcal{f}$ is not injective



Are there any mistakes in these writings?






examples-counterexamples







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Need some help to check whether my understanding of a subject is right or not.



    $1$. Example of injective function.



    $mathcal{f}$: $mathbb{N}$ $longrightarrow$ $mathbb{N}$



    $x$ $longmapsto$ $2x$



    A = {$1,2,3,4$}, B = {$2,4,5,7$}, (A $cap$ B) = {$2,4$}



    $mathcal{f}$(A) = {$2,4,6,8$}, $space$ $mathcal{f}$(B) = {$4,8,10,14$}



    $mathcal{f}$(A $cap$ B) = $mathcal{f}$(A) $cap$ $mathcal{f}$(B) = {$4,8$} $Longrightarrow$ $mathcal{f}$ is injective



    $2$. Example of a function that is not injective



    $mathcal{f}$: $mathbb{R}$ $longrightarrow$ $mathbb{R}$



    $x$ $longmapsto$ $x^2$



    A = {$2,4,6,8$}, B = {$-4,-8,-12,-16$}, (A $cap$ B) = $lbrace$ $emptyset$ $rbrace$



    $mathcal{f}$(A) = {$4,16,36,64$}, $space$ $mathcal{f}$(B) = {$16,64,144,256$}



    $mathcal{f}$(A) $cap$ $mathcal{f}$(B) = {$16,64$} $neq$ $mathcal{f}$(A $cap$ B) = $lbrace$ $emptyset$ $rbrace$ $Longrightarrow$ $mathcal{f}$ is not injective



    Are there any mistakes in these writings?






    examples-counterexamples







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Need some help to check whether my understanding of a subject is right or not.



      $1$. Example of injective function.



      $mathcal{f}$: $mathbb{N}$ $longrightarrow$ $mathbb{N}$



      $x$ $longmapsto$ $2x$



      A = {$1,2,3,4$}, B = {$2,4,5,7$}, (A $cap$ B) = {$2,4$}



      $mathcal{f}$(A) = {$2,4,6,8$}, $space$ $mathcal{f}$(B) = {$4,8,10,14$}



      $mathcal{f}$(A $cap$ B) = $mathcal{f}$(A) $cap$ $mathcal{f}$(B) = {$4,8$} $Longrightarrow$ $mathcal{f}$ is injective



      $2$. Example of a function that is not injective



      $mathcal{f}$: $mathbb{R}$ $longrightarrow$ $mathbb{R}$



      $x$ $longmapsto$ $x^2$



      A = {$2,4,6,8$}, B = {$-4,-8,-12,-16$}, (A $cap$ B) = $lbrace$ $emptyset$ $rbrace$



      $mathcal{f}$(A) = {$4,16,36,64$}, $space$ $mathcal{f}$(B) = {$16,64,144,256$}



      $mathcal{f}$(A) $cap$ $mathcal{f}$(B) = {$16,64$} $neq$ $mathcal{f}$(A $cap$ B) = $lbrace$ $emptyset$ $rbrace$ $Longrightarrow$ $mathcal{f}$ is not injective



      Are there any mistakes in these writings?






      examples-counterexamples







      share|cite|improve this question











      $endgroup$




      Need some help to check whether my understanding of a subject is right or not.



      $1$. Example of injective function.



      $mathcal{f}$: $mathbb{N}$ $longrightarrow$ $mathbb{N}$



      $x$ $longmapsto$ $2x$



      A = {$1,2,3,4$}, B = {$2,4,5,7$}, (A $cap$ B) = {$2,4$}



      $mathcal{f}$(A) = {$2,4,6,8$}, $space$ $mathcal{f}$(B) = {$4,8,10,14$}



      $mathcal{f}$(A $cap$ B) = $mathcal{f}$(A) $cap$ $mathcal{f}$(B) = {$4,8$} $Longrightarrow$ $mathcal{f}$ is injective



      $2$. Example of a function that is not injective



      $mathcal{f}$: $mathbb{R}$ $longrightarrow$ $mathbb{R}$



      $x$ $longmapsto$ $x^2$



      A = {$2,4,6,8$}, B = {$-4,-8,-12,-16$}, (A $cap$ B) = $lbrace$ $emptyset$ $rbrace$



      $mathcal{f}$(A) = {$4,16,36,64$}, $space$ $mathcal{f}$(B) = {$16,64,144,256$}



      $mathcal{f}$(A) $cap$ $mathcal{f}$(B) = {$16,64$} $neq$ $mathcal{f}$(A $cap$ B) = $lbrace$ $emptyset$ $rbrace$ $Longrightarrow$ $mathcal{f}$ is not injective



      Are there any mistakes in these writings?






      examples-counterexamples




      examples-counterexamples






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 5 '18 at 19:38







      Egor Epishin

















      asked Dec 5 '18 at 13:51









      Egor EpishinEgor Epishin

      54




      54






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Yes, there is a mistake, and a big one: you cannot fix two sets $A$ and $B$ and claim that$$f(Acap B)=f(A)cap f(B)implies ftext{ injective.}$$That equality holds for those two specific set, but that fact, by itslef, is not enough to assure that $f$ is injective.



          The rest looks correct.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            however, if it holds for all subsets, or at least the singleton subsets, it should work out.
            $endgroup$
            – Enkidu
            Dec 5 '18 at 14:00










          • $begingroup$
            @Enkidu Indeed, but the OP only claimed that it holds for two specific sets.
            $endgroup$
            – José Carlos Santos
            Dec 5 '18 at 14:01










          • $begingroup$
            Thank you for the clarification. So, if the statement (from the title of the post) holds for all subsets of the domain of a function then we can say that this function is injective. While we only need at least two arbitrarily subsets of the domain for which the statement doesn't hold to state that the function is not injective. Right?
            $endgroup$
            – Egor Epishin
            Dec 5 '18 at 14:54










          • $begingroup$
            Completely right!
            $endgroup$
            – José Carlos Santos
            Dec 5 '18 at 14:56



















          0












          $begingroup$

          Here's another mistake: If $A = {2,4,6,8}, B = {-4,-8,-12,-16}$ then $A cap B ne { emptyset }$.



          $A cap B = emptyset$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Maybe I misunderstood you. You are saying that there is a mistake in the following line "If A={2,4,6,8}, B={−4,−8,−12,−16} then A∩B ≠ {∅}". However, in my post I write that A∩B = ∅.
            $endgroup$
            – Egor Epishin
            Dec 5 '18 at 15:03










          • $begingroup$
            No, in your post you wrote $A cap B = {emptyset}$. Be careful: ${emptyset} ne emptyset$
            $endgroup$
            – jjagmath
            Dec 5 '18 at 15:07












          • $begingroup$
            I see. Thank you!
            $endgroup$
            – Egor Epishin
            Dec 5 '18 at 15:13











          Your Answer





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          2 Answers
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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Yes, there is a mistake, and a big one: you cannot fix two sets $A$ and $B$ and claim that$$f(Acap B)=f(A)cap f(B)implies ftext{ injective.}$$That equality holds for those two specific set, but that fact, by itslef, is not enough to assure that $f$ is injective.



          The rest looks correct.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            however, if it holds for all subsets, or at least the singleton subsets, it should work out.
            $endgroup$
            – Enkidu
            Dec 5 '18 at 14:00










          • $begingroup$
            @Enkidu Indeed, but the OP only claimed that it holds for two specific sets.
            $endgroup$
            – José Carlos Santos
            Dec 5 '18 at 14:01










          • $begingroup$
            Thank you for the clarification. So, if the statement (from the title of the post) holds for all subsets of the domain of a function then we can say that this function is injective. While we only need at least two arbitrarily subsets of the domain for which the statement doesn't hold to state that the function is not injective. Right?
            $endgroup$
            – Egor Epishin
            Dec 5 '18 at 14:54










          • $begingroup$
            Completely right!
            $endgroup$
            – José Carlos Santos
            Dec 5 '18 at 14:56
















          1












          $begingroup$

          Yes, there is a mistake, and a big one: you cannot fix two sets $A$ and $B$ and claim that$$f(Acap B)=f(A)cap f(B)implies ftext{ injective.}$$That equality holds for those two specific set, but that fact, by itslef, is not enough to assure that $f$ is injective.



          The rest looks correct.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            however, if it holds for all subsets, or at least the singleton subsets, it should work out.
            $endgroup$
            – Enkidu
            Dec 5 '18 at 14:00










          • $begingroup$
            @Enkidu Indeed, but the OP only claimed that it holds for two specific sets.
            $endgroup$
            – José Carlos Santos
            Dec 5 '18 at 14:01










          • $begingroup$
            Thank you for the clarification. So, if the statement (from the title of the post) holds for all subsets of the domain of a function then we can say that this function is injective. While we only need at least two arbitrarily subsets of the domain for which the statement doesn't hold to state that the function is not injective. Right?
            $endgroup$
            – Egor Epishin
            Dec 5 '18 at 14:54










          • $begingroup$
            Completely right!
            $endgroup$
            – José Carlos Santos
            Dec 5 '18 at 14:56














          1












          1








          1





          $begingroup$

          Yes, there is a mistake, and a big one: you cannot fix two sets $A$ and $B$ and claim that$$f(Acap B)=f(A)cap f(B)implies ftext{ injective.}$$That equality holds for those two specific set, but that fact, by itslef, is not enough to assure that $f$ is injective.



          The rest looks correct.






          share|cite|improve this answer









          $endgroup$



          Yes, there is a mistake, and a big one: you cannot fix two sets $A$ and $B$ and claim that$$f(Acap B)=f(A)cap f(B)implies ftext{ injective.}$$That equality holds for those two specific set, but that fact, by itslef, is not enough to assure that $f$ is injective.



          The rest looks correct.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 13:57









          José Carlos SantosJosé Carlos Santos

          156k22125227




          156k22125227












          • $begingroup$
            however, if it holds for all subsets, or at least the singleton subsets, it should work out.
            $endgroup$
            – Enkidu
            Dec 5 '18 at 14:00










          • $begingroup$
            @Enkidu Indeed, but the OP only claimed that it holds for two specific sets.
            $endgroup$
            – José Carlos Santos
            Dec 5 '18 at 14:01










          • $begingroup$
            Thank you for the clarification. So, if the statement (from the title of the post) holds for all subsets of the domain of a function then we can say that this function is injective. While we only need at least two arbitrarily subsets of the domain for which the statement doesn't hold to state that the function is not injective. Right?
            $endgroup$
            – Egor Epishin
            Dec 5 '18 at 14:54










          • $begingroup$
            Completely right!
            $endgroup$
            – José Carlos Santos
            Dec 5 '18 at 14:56


















          • $begingroup$
            however, if it holds for all subsets, or at least the singleton subsets, it should work out.
            $endgroup$
            – Enkidu
            Dec 5 '18 at 14:00










          • $begingroup$
            @Enkidu Indeed, but the OP only claimed that it holds for two specific sets.
            $endgroup$
            – José Carlos Santos
            Dec 5 '18 at 14:01










          • $begingroup$
            Thank you for the clarification. So, if the statement (from the title of the post) holds for all subsets of the domain of a function then we can say that this function is injective. While we only need at least two arbitrarily subsets of the domain for which the statement doesn't hold to state that the function is not injective. Right?
            $endgroup$
            – Egor Epishin
            Dec 5 '18 at 14:54










          • $begingroup$
            Completely right!
            $endgroup$
            – José Carlos Santos
            Dec 5 '18 at 14:56
















          $begingroup$
          however, if it holds for all subsets, or at least the singleton subsets, it should work out.
          $endgroup$
          – Enkidu
          Dec 5 '18 at 14:00




          $begingroup$
          however, if it holds for all subsets, or at least the singleton subsets, it should work out.
          $endgroup$
          – Enkidu
          Dec 5 '18 at 14:00












          $begingroup$
          @Enkidu Indeed, but the OP only claimed that it holds for two specific sets.
          $endgroup$
          – José Carlos Santos
          Dec 5 '18 at 14:01




          $begingroup$
          @Enkidu Indeed, but the OP only claimed that it holds for two specific sets.
          $endgroup$
          – José Carlos Santos
          Dec 5 '18 at 14:01












          $begingroup$
          Thank you for the clarification. So, if the statement (from the title of the post) holds for all subsets of the domain of a function then we can say that this function is injective. While we only need at least two arbitrarily subsets of the domain for which the statement doesn't hold to state that the function is not injective. Right?
          $endgroup$
          – Egor Epishin
          Dec 5 '18 at 14:54




          $begingroup$
          Thank you for the clarification. So, if the statement (from the title of the post) holds for all subsets of the domain of a function then we can say that this function is injective. While we only need at least two arbitrarily subsets of the domain for which the statement doesn't hold to state that the function is not injective. Right?
          $endgroup$
          – Egor Epishin
          Dec 5 '18 at 14:54












          $begingroup$
          Completely right!
          $endgroup$
          – José Carlos Santos
          Dec 5 '18 at 14:56




          $begingroup$
          Completely right!
          $endgroup$
          – José Carlos Santos
          Dec 5 '18 at 14:56











          0












          $begingroup$

          Here's another mistake: If $A = {2,4,6,8}, B = {-4,-8,-12,-16}$ then $A cap B ne { emptyset }$.



          $A cap B = emptyset$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Maybe I misunderstood you. You are saying that there is a mistake in the following line "If A={2,4,6,8}, B={−4,−8,−12,−16} then A∩B ≠ {∅}". However, in my post I write that A∩B = ∅.
            $endgroup$
            – Egor Epishin
            Dec 5 '18 at 15:03










          • $begingroup$
            No, in your post you wrote $A cap B = {emptyset}$. Be careful: ${emptyset} ne emptyset$
            $endgroup$
            – jjagmath
            Dec 5 '18 at 15:07












          • $begingroup$
            I see. Thank you!
            $endgroup$
            – Egor Epishin
            Dec 5 '18 at 15:13
















          0












          $begingroup$

          Here's another mistake: If $A = {2,4,6,8}, B = {-4,-8,-12,-16}$ then $A cap B ne { emptyset }$.



          $A cap B = emptyset$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Maybe I misunderstood you. You are saying that there is a mistake in the following line "If A={2,4,6,8}, B={−4,−8,−12,−16} then A∩B ≠ {∅}". However, in my post I write that A∩B = ∅.
            $endgroup$
            – Egor Epishin
            Dec 5 '18 at 15:03










          • $begingroup$
            No, in your post you wrote $A cap B = {emptyset}$. Be careful: ${emptyset} ne emptyset$
            $endgroup$
            – jjagmath
            Dec 5 '18 at 15:07












          • $begingroup$
            I see. Thank you!
            $endgroup$
            – Egor Epishin
            Dec 5 '18 at 15:13














          0












          0








          0





          $begingroup$

          Here's another mistake: If $A = {2,4,6,8}, B = {-4,-8,-12,-16}$ then $A cap B ne { emptyset }$.



          $A cap B = emptyset$






          share|cite|improve this answer









          $endgroup$



          Here's another mistake: If $A = {2,4,6,8}, B = {-4,-8,-12,-16}$ then $A cap B ne { emptyset }$.



          $A cap B = emptyset$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 14:41









          jjagmathjjagmath

          2047




          2047












          • $begingroup$
            Maybe I misunderstood you. You are saying that there is a mistake in the following line "If A={2,4,6,8}, B={−4,−8,−12,−16} then A∩B ≠ {∅}". However, in my post I write that A∩B = ∅.
            $endgroup$
            – Egor Epishin
            Dec 5 '18 at 15:03










          • $begingroup$
            No, in your post you wrote $A cap B = {emptyset}$. Be careful: ${emptyset} ne emptyset$
            $endgroup$
            – jjagmath
            Dec 5 '18 at 15:07












          • $begingroup$
            I see. Thank you!
            $endgroup$
            – Egor Epishin
            Dec 5 '18 at 15:13


















          • $begingroup$
            Maybe I misunderstood you. You are saying that there is a mistake in the following line "If A={2,4,6,8}, B={−4,−8,−12,−16} then A∩B ≠ {∅}". However, in my post I write that A∩B = ∅.
            $endgroup$
            – Egor Epishin
            Dec 5 '18 at 15:03










          • $begingroup$
            No, in your post you wrote $A cap B = {emptyset}$. Be careful: ${emptyset} ne emptyset$
            $endgroup$
            – jjagmath
            Dec 5 '18 at 15:07












          • $begingroup$
            I see. Thank you!
            $endgroup$
            – Egor Epishin
            Dec 5 '18 at 15:13
















          $begingroup$
          Maybe I misunderstood you. You are saying that there is a mistake in the following line "If A={2,4,6,8}, B={−4,−8,−12,−16} then A∩B ≠ {∅}". However, in my post I write that A∩B = ∅.
          $endgroup$
          – Egor Epishin
          Dec 5 '18 at 15:03




          $begingroup$
          Maybe I misunderstood you. You are saying that there is a mistake in the following line "If A={2,4,6,8}, B={−4,−8,−12,−16} then A∩B ≠ {∅}". However, in my post I write that A∩B = ∅.
          $endgroup$
          – Egor Epishin
          Dec 5 '18 at 15:03












          $begingroup$
          No, in your post you wrote $A cap B = {emptyset}$. Be careful: ${emptyset} ne emptyset$
          $endgroup$
          – jjagmath
          Dec 5 '18 at 15:07






          $begingroup$
          No, in your post you wrote $A cap B = {emptyset}$. Be careful: ${emptyset} ne emptyset$
          $endgroup$
          – jjagmath
          Dec 5 '18 at 15:07














          $begingroup$
          I see. Thank you!
          $endgroup$
          – Egor Epishin
          Dec 5 '18 at 15:13




          $begingroup$
          I see. Thank you!
          $endgroup$
          – Egor Epishin
          Dec 5 '18 at 15:13


















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          Province de Neuquén

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          Province de Neuquén Provincia del Neuquén .mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Héraldique Drapeau Localisation de la province de Neuquén Administration Pays   Argentine Capitale Neuquén Gouverneur Omar Gutiérrez ISO 3166-2 AR-Q Démographie Gentilé Neuquino/a Population 550 344  hab. (2010) Densité 5,8  hab./km 2 Géographie Superficie 94 078  km 2 Liens Site web http://www.neuquen.gov.ar modifier   La province de Neuquén est une province d'Argentine, qui se trouve à l'extrémité nord-ouest de la Patagonie argentine. Avec une superficie de 94 378  km 2 , elle est la plus petite province des régions australes du pays. Sa géographie s'abîme sur l'imposante Cordillère des Andes. Sommaire 1 Histoire 1.1 Avant la découverte par les Espagnols 1.2 ...
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          Journaliste

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          Pour les articles homonymes, voir Journaliste (homonymie). Journaliste Journalistes Appellation Journaliste Secteur d'activité Média - Communication Professions voisines Journaliste sportif - Reporter-photographe - Éditorialiste Codes CITP 2451, 2642 IDEO (France) 10197 ROME (France) E1106 modifier - modifier le code - modifier Wikidata Un journaliste est une personne qui recueille, écrit ou distribue des informations. L'activité d'un journaliste est le journalisme. Un journaliste peut travailler sur des sujets généraux ou se spécialiser dans certaines questions. La plupart des journalistes ont tendance à se spécialiser et, en collaborant avec d'autres journalistes, rédigent un article ou mettent en forme un reportage qui couvre l'actualité. Par exemple, un journaliste sportif couvre les nouvelles dans le monde du sport, mais ce journaliste peut aussi faire partie d'un journal qui couvre d'autres su...
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