{"id":3312,"date":"2012-04-01T02:22:00","date_gmt":"2012-03-31T17:22:00","guid":{"rendered":"https:\/\/mathematicians.korea.ac.kr\/ywkim\/?p=3312"},"modified":"2021-09-01T20:28:15","modified_gmt":"2021-09-01T11:28:15","slug":"2012-%ec%a7%91%ed%95%a9%eb%a1%a0-%eb%ac%b8%ec%a0%9c-%ed%8e%98%ec%9d%b4%ec%a7%80","status":"publish","type":"post","link":"https:\/\/mathematicians.korea.ac.kr\/ywkim\/2012\/04\/01\/2012-%ec%a7%91%ed%95%a9%eb%a1%a0-%eb%ac%b8%ec%a0%9c-%ed%8e%98%ec%9d%b4%ec%a7%80\/","title":{"rendered":"2012 \uc9d1\ud569\ub860 \ubb38\uc81c \ud398\uc774\uc9c0"},"content":{"rendered":"
\n

22, March.<\/h2>\n
\n
    \n
  1. Check that $ f^{-1}:B→ A$ ( $ a=f^{-1}(b)$ \u21d4 $ b=f(a)$ ) is well-defined.<\/li>\n
  2. ”Lemma 2.1” proof.<\/li>\n
  3. ”Exercise #5” (p.21), and consider why this should be 1-1, onto.<\/li>\n
  4. Check these. \u2460 $ f:A→ B$ : 1-1 \u21d4 \\(\\forall g,h:Z\\to B\\) ( $ fˆ g = fˆ h$ \u21d2 $ g=h$ ) and \u2461 $ f:A→ B$ : onto \u21d4 \\(\\forall g,h:B\\to Z\\) ( $ gˆ f = hˆ f$ \u21d2 $ g=h$ ).<\/li>\n
  5. ”Exercise #2” (p.20), and consider counterexample to know why.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n
    \n

    20, March.<\/h2>\n
    \n
      \n
    1. Prove that $ f∪(A_0× B)$ . [restriction part]<\/li>\n<\/ol>\n

      \ud639\uc2dc, $ f∩ A_0× B = $ restriction of $ f $ w.r.t. $ A_0 $ \uc744 \ubb3b\ub294 \ubb38\uc81c\uc778\uac00\uc694? -\uc774 \ub9d0\uc774 \ub9de\uc544\uc694. <\/p>\n

        \n
      1. Check $ (gˆ f)$ is a function. [composite fn part]<\/li>\n
      2. Choose and translate one section of Naive Set Theory(after sec.4, Halmos) or Topology(textbook, Munkres). [submit at the end of march]<\/li>\n<\/ol>\n

        \uba54\uc77c\ub85c \uc81c\ucd9c\ud558\ub294 \uac74\uac00\uc694? \ud559\uad50 \uba54\uc77c\ub85c \ubcf4\ub0c8\uc2a4\u3142\ub2c8\ub2e4 <\/p>\n<\/div>\n<\/div>\n

        \n

        15, March.<\/h2>\n
        \n
          \n
        1. Define distrubutive, associative, communicative law and complement of union\/intersection in general sense.<\/li>\n
        2. Prove that if $ a=b$ , $ (a,b)$ is singleton.<\/li>\n<\/ol>\n<\/div>\n
          \n

          Solution<\/h3>\n
          \n

          2: By the definition of oredered pairs, <\/p>\n

          $ (a,b)=\\{\\{a\\},\\{a,b\\}\\}$ <\/p>\n

          If $ a=b$ , $ (a,a)=\\{\\{a\\},\\{a,a\\}\\}$ and $ \\{\\{a\\},\\{a,a\\}\\}=\\{\\{a\\},\\{a\\}\\}$ . <\/p>\n

          (\u2235 If two elements is same, set only express one element. Therefore $ \\{a\\}=\\{a,b\\}$ if $ a=b$ ) <\/p>\n

          By the same manner, $ \\{\\{a\\},\\{a\\}\\}=\\{\\{a\\}\\}$ . <\/p>\n

          Therefore, if $ a=b$ , $ (a,b)$ \u21d2 $ \\{\\{a\\}\\}$ . The proof is complete. <\/p>\n

          -2012. 3. 15 2006140631 \uc774\uc2b9\ud6c4 <\/p>\n

            \n
          1. “set only express one element.” \uc774 \uba85\uc81c\uc5d0 \ub300\ud55c \uc99d\uba85\uc774 \ud544\uc694\ud558\uc9c0 \uc54a\ub098 \uc0dd\uac01\ud574 \ubd24\uc2b5\ub2c8\ub2e4.<\/li>\n<\/ol>\n

            a\ub97c \ub450\uac1c \uac00\uc9c4 \uc9d1\ud569\uacfc a\ub97c 1\uac1c \uac00\uc9c4 \uc9d1\ud569\uc744 \uac19\ub2e4\uace0 \ubc1b\uc544\ub4e4\uc77c \uc218\uac00 \uc5c6\uc5c8\uc2b5\ub2c8\ub2e4. \uc194\uc9c1\ud788 ‘\ud560\ubaa8\uc2a4\uac00 \uc774 \ubb38\uc81c \uc804\uae4c\uc9c0 \uc81c\uc2dc\ud55c \uacf5\ub9ac\ub4e4\ub9cc\uc744 \uac00\uc9c0\uace0\ub3c4 a\ub97c 1\uac1c \uac00\uc9c4 \uc9d1\ud569\uc774 \uc815\ub9d0 \uc874\uc7ac\ud558\ub294 \uc9c0 \ub2f5\ud560 \uc218 \uc788\uc744\uae4c’\ub77c\ub294 \uc0dd\uac01\ub3c4 \ub4e4\uc5c8\uc2b5\ub2c8\ub2e4. \uc77c\ub2e8 \uc874\uc7ac\ud574\uc57c {a, a} = {a}\ub780 \uba85\uc81c\ub3c4 \uac00\ub2a5\ud558\uc796\uc544\uc694. \ubb50 \ucc45\uc5d0\uc11c\ub294 {a, a}\ub97c {a}\ub85c \ud45c\uae30\ud558\uc790\uace0 \uc57d\uc18d\ud558\uc9c0\ub9cc \ucca0\uc218\uac00 \ub2e4\ub978 \uc774\ub984\uc744 \uac16\ub294\ub2e4\uace0,\uc608\ub97c \ub4e4\uba74 \uc601\ud76c, \uc9c4\uc9dc \uc601\ud76c\uac00 \ub418\ub294 \uac74 \uc544\ub2cc \uac83 \uac19\uad6c\uc694^^ \uace0\ubbfc\ud558\ub2e4\uac00 \uc774\ub7f0 \uc2dd\uc758 \ub17c\ub9ac\ub97c \ud3b4\ubd24\uc2b5\ub2c8\ub2e4. <\/p>\n

            Suppose a set, A exists. Then, by the axiom of pairs, we know a set, {A, A} also exists. This time, suppose a set containing A only exists and let’s denote it as {A}. Now, I claim {A, A} = {A}. <- a proof on it can be easily given through usual ways. Therefore, the set {A} really exists and that is {A, A} <\/p>\n

            \uc870\uae08 \ucc1c\ucc1c\ud55c \uac83\uc740 \uc774\ubbf8 \uc874\uc7ac\ud558\ub294 P\uac00 \uac00\uc0c1\uc758 Q\uc640 \uac19\ub2e4\uace0 \ud574\uc11c P = Q\uc774\uace0 \ub530\ub77c\uc11c Q\ub294 \uc874\uc7ac\ud55c\ub2e4\uace0\ud55c \ubd80\ubd84\uc785\ub2c8\ub2e4. \uacc4\uc18d \uacf1\uc539\ub2e4\ubcf4\ub2c8 \uadf8\ub7f0 \uac83 \uac19\uae30\uace0 \ud558\ub124\uc694^^;; \uadf8\ub7f0\ub370 \uc904\ubc14\uafc8\uc740 \uc5b4\ub5bb\uac8c \ud558\uc8e0, \uc8c4\uc1a1. <\/p>\n

            \uc9c8\ubb38 : \ud560\ubaa8\uc2a4\ub97c \ub530\ub978\ub2e4\uba74 “set only express one element.” \uc774 \uba85\uc81c\uc758 \uc99d\uba85\uc774 \ud544\uc694\ud55c\uac00\uc694? \uadfc\ub370 \uc774 \ubb38\uc7a5\uc774 \ud560\ubaa8\uc2a4 \ucc45 \uc5b4\ub514\uc5d0 \uc788\ub294 \uac74\uac00\uc694? \uba87\ucabd, \uba87\uc9f8\uc904? \ud560\ubaa8\uc2a4 \ucc45\uc5d0 \ub098\uc624\uc9c4 \uc54a\uace0 \uc704\uc5d0 \uc774\uc2b9\ud6c4\uc528\uac00 \uc99d\uba85\uc5d0\uc11c \uc0ac\uc6a9\ud55c \uba85\uc81c\uc785\ub2c8\ub2e4. <\/p>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

            22, March. Check that $ f^{-1}:B→ A$ ( $ a=f^{-1}(b)$ \u21d4 $ b=f(a)$ ) is well-defined. ”Lemma 2.1” proof. ”Exercise #5” (p.21), and consider why this should be 1-1, onto. Check these. \u2460 $ f:A→ B$ : 1-1 \u21d4 \\(\\forall g,h:Z\\to B\\) ( $ fˆ g = fˆ h$ \u21d2 $ g=h$ ) and \u2461 … \ub354 \uc77d\uae30<\/a><\/p>\n","protected":false},"author":7,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"ngg_post_thumbnail":0,"footnotes":""},"categories":[11],"tags":[],"class_list":["post-3312","post","type-post","status-publish","format-standard","hentry","category-lectures"],"distributor_meta":false,"distributor_terms":false,"distributor_media":false,"distributor_original_site_name":"\uae40\uc601\uc6b1","distributor_original_site_url":"https:\/\/mathematicians.korea.ac.kr\/ywkim","push-errors":false,"_links":{"self":[{"href":"https:\/\/mathematicians.korea.ac.kr\/ywkim\/wp-json\/wp\/v2\/posts\/3312","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mathematicians.korea.ac.kr\/ywkim\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathematicians.korea.ac.kr\/ywkim\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathematicians.korea.ac.kr\/ywkim\/wp-json\/wp\/v2\/users\/7"}],"replies":[{"embeddable":true,"href":"https:\/\/mathematicians.korea.ac.kr\/ywkim\/wp-json\/wp\/v2\/comments?post=3312"}],"version-history":[{"count":1,"href":"https:\/\/mathematicians.korea.ac.kr\/ywkim\/wp-json\/wp\/v2\/posts\/3312\/revisions"}],"predecessor-version":[{"id":3313,"href":"https:\/\/mathematicians.korea.ac.kr\/ywkim\/wp-json\/wp\/v2\/posts\/3312\/revisions\/3313"}],"wp:attachment":[{"href":"https:\/\/mathematicians.korea.ac.kr\/ywkim\/wp-json\/wp\/v2\/media?parent=3312"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathematicians.korea.ac.kr\/ywkim\/wp-json\/wp\/v2\/categories?post=3312"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathematicians.korea.ac.kr\/ywkim\/wp-json\/wp\/v2\/tags?post=3312"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}