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Real Analysis I
äÁè·ÃÒºÇèÒ¾ÔÊÙ¨¹ìà຺¹Õé¨ÐâÍà¤ÃÖà»ÅèÒ¤ÃѺ ú¡Ç¹¼ÙéÃÙé´éǹФÃѺ
¨§¾ÔÊÙ¨¹ìÇèÒ ¶éÒ $S,T$ à»ç¹à«µ·ÕèÁբͺࢵ ààÅéǨÐä´éÇèÒ $\sup(S\cup T)=\max\left\{\,\sup(S),\sup(T)\right\} $ Solution $Let$ $a\in S\cup T$ $we$ $have$ $a\in S$ $or$ $a\in T$ $.if$ $a\in S$ $then$ $a \le \sup(S)$ $and$ $if$ $a\in T$ $then$ $a\le \sup(T)$ $Hence,$ $a\le \max\left\{\,\sup(S),\sup(T)\right\} $ $So$ $\max\left\{\,\sup(S),\sup(T)\right\}$ $is$ $the$ $upper$ $boundary$ $of$ $S\cup T.$ $Let$ $p$ $be$ $another$ $upper$ $boundary$ $of$ $S\cup T$ $such$ $that$ $p<\max\left\{\,\sup(S),\sup(T)\right\}$ $because$ $p$ $is$ $the$ $upper$ $boundary.$ $Thus,$ $p\ge \sup(S)$ $and$ $p\ge \sup(T)$ $which$ $implies$ $\max\left\{\,\sup(S),\sup(T)\right\}> p\ge \max\left\{\,\sup(S),\sup(T)\right\}$ $contradiction.$ $So$ $\max\left\{\,\sup(S),\sup(T)\right\}=\sup(S\cup T)$
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Vouloir c'est pouvoir 10 Á¡ÃÒ¤Á 2016 22:18 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 2 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ ¨Ù¡Ñ´àËÅÕ§ |
#2
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¡çâÍहФÃѺ áµè¢Íá¹Ð¹ÓµÃ§¡ÒÃà¢Õ¹¹Ô´Ë¹èÍ 1) upper boundary à¢ÒäÁèàÃÕ¡¡Ñ¹¤ÃѺ à¢ÒàÃÕ¡ upper bondà©Âæ
2) "the" upper bond ãªéäÁèä´é¹Ð¤ÃѺ ¶éÒàÃÒäÁèä´éÁÕupper bondµÑÇà´ÕÂÇ! µéͧà»ÅÕè¹à»ç¹ an ¤ÃѺ ÍÒ¨¨Ð´ÙäÁèÊÓ¤Ñ áµèÁѹ·ÓãËé¤ÇÒÁËÁÒÂà»ÅÕè¹仫Ö觡ÒÃà¢Õ¹¾ÔÊÙ¨¹ì·Õè´ÕäÁè¤ÇÃ·Ó the ãªéä´é¡Ñº supremum ¤ÃѺ à¾ÃÒÐàÃÒä´é¾ÔÊÙ¨¹ìÁÒáÅéÇÇèÒ supremum ¢Í§áµèÅÐ૵¹Ñé¹ unique
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Mathematics, rightly viewed possesses not only truth, but supreme beauty. B.R. |
#3
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First, we note that all the terms $\sup(S)$, $\sup(T)$ and $\sup(S\cup T)$ exist. We now divide the proof into two parts.
Part 1: We show that $\sup(S)\le\sup(S\cup T)$ and $\sup(T)\le\sup(S\cup T)$. (Hence $\max\{\sup(S),\sup(T)\}\le\sup(S\cup T)$.) Since $S\subset S\cup T$, we have $\sup(S\cup T)$ is an upper bound of $S$. Since $\sup(S)$ is the least upper bound of $S$, we have $\sup(S)\le\sup(S\cup T)$. Similarly, we can prove that $\sup(T)\le\sup(S\cup T)$. Part 2: We show that $\sup(S\cup T)\le\max\{\sup(S),\sup(T)\}$. To prove this statement, we show that $\max\{\sup(S),\sup(T)\}$ is an upper bound of $S\cup T$. Let $x\in S\cup T$. This implies that $x\in S$ or $x\in T$. If $x\in S$, then $x\le\sup S\le \max\{\sup(S),\sup(T)\}$. If $x\in T$, then $x\le\sup T\le \max\{\sup(S),\sup(T)\}$. Since $\sup(S\cup T)$ is the least upper bound of $S\cup T$, we have $\sup(S\cup T)\le\max\{\sup(S),\sup(T)\}$. |
#4
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ÊÁѼÁàÃÕ¹ ¡çµéͧ¡૵¨Ó¹Ç¹àµçÁ ãªé·´Êͺà¹×éÍËÒ·ÕèÍèÒ¹ÇèÒà»ç¹¨ÃÔ§äËÁ
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