{"id":515,"date":"2016-11-02T14:46:21","date_gmt":"2016-11-02T18:46:21","guid":{"rendered":"http:\/\/blog.richmond.edu\/math320\/?p=515"},"modified":"2017-08-22T17:00:47","modified_gmt":"2017-08-22T21:00:47","slug":"a-short-words-proof-of-n-i-p","status":"publish","type":"post","link":"https:\/\/blog.richmond.edu\/math320\/2016\/11\/02\/a-short-words-proof-of-n-i-p\/","title":{"rendered":"A Short Words Proof of N.I.P."},"content":{"rendered":"

The Challenge:<\/strong>\u00a0<\/span>Write a monosyllabic version of a proof we have studied this semester.<\/p>\n

Accepting the Challenge:<\/strong><\/span>\u00a0My strategy was to choose a proof that was visually easy to understand and hope that I can convey the basic idea of the proof to the reader as an image. The proof I chose was the Nested Interval Property. Below is a more mathematically formal version of the proof. Even further below, you can find my simplified version.<\/p>\n

<\/p>\n

Nested Interval Property:<\/strong><\/span>
\nDefinition: <\/strong><\/span>For all \"n\in, let \"I_n and let \"I_{n+1}\subseteq. Then \"\cap^\infty_{n=1}I_n\not=\emptyset\".<\/p>\n

Proof:<\/strong><\/span> For all \"n\in, let \"I_n and let \"I_{n+1}\subseteq. Let \"A. Consider \"b_1\". Since \"\forall\" \"n\in\mathbb{N}\", \"I_n\subseteq, it follows that \"a_1\le. Thus, for all \"n\in, \"a_n\le, so \"b_1\" is an upper bound for \"A\". Since A<\/em>\u00a0is bounded above, the axiom of completeness guarantees the existence of \"x=\sup(A)\". For all n<\/em>, \"b_n\" is an upper bound for A<\/em>\u00a0by nestedness. But \"x\le, since \"x. Therefore, \"a_n, which implies \"x\in. Finally, \"x\in.\"\Box\"<\/p>\n

Here we go…<\/strong><\/span>
\nWhat It Is:<\/strong><\/span> Let \"I_n\" be a closed span of reals such that \"a_n\" is the left most part, \"b_n\" is the right most part, and \"a_n\" is less than \"b_n\". All the sets, \"I_n\", can be made out by their n<\/em>, where n<\/em>\u00a0is a whole count: 1,2,3,…<\/em>. As n<\/em>\u00a0gets big, if each \"I_n\" lies in the past set, then the core of all the sets is not void.\\\\<\/p>\n

Proof:<\/strong><\/span> First, let \"I_n\" be a closed span of reals, where \"a_n\" is the left most part, b<\/em> is the right most part, and \"a_n\" is less than \"b_n\". And, as\u00a0n<\/em> gets big, each \"I_n\" lies in the past set. Let A<\/em> be the set of left most points, \"a_n\". Let \"b_1\" be the right most part of the first set, \"I_1\". Since, as n<\/em>\u00a0gets big, \"I_n\" lies in the past set, we can say that \"b_1\" is more than all \"b_n\", which we know is more than \"a_n\". Thus, \"a_n\" is less than \"b_1\".<\/p>\n

This means that \"b_1\" is more than all parts in the set A<\/em>\u00a0and is a bound for A<\/em>. Now, we use a math truth that says a set with a bound that is more than all the parts of the set will have a bound that is less than or the same as all the bounds of A<\/em>. So, the set of bounds for A<\/em>\u00a0will have a least part. We will call this the “small bound”.<\/p>\n

From this truth we can say that there is a small bound of A<\/em>. Let x<\/em>\u00a0be the small bound. Then x<\/em>\u00a0is less than or the same as all \"b_n\". But, we know \"b_n\" is more than all \"a_n\", so it too is a bound for A<\/em>. So,\u00a0x<\/em> is less than or the same as all \"b_n\" and more than or the same as all \"a_n\". Thus, \"x\" must be in the set \"I_n\". Then x<\/em>\u00a0is in the core all the sets \"I_n\" and so \"I_n\" can’t be void.\"\Box\"<\/p>\n

\"Image1[4]<\/a>
\nArt 1: The core of the sets \"I_n\" with x<\/em>\u00a0in the core. The sets \"I_1\", \"I_2\", \"I_3\", \"I_4\" are shown so as to give some sense of what the core of all the sets is.<\/p>\n

\"Image2[1]<\/a>
\nArt 2: Here, the sets \"I_1\", \"I_2\", \"I_3\", \"I_4\" are shown so as to give some sense of what the left most parts and right most parts are in terms of the sets.<\/p>\n

Post Challenge:<\/strong><\/span>\u00a0Einstein phrased it perfectly when he said, “If you can’t explain it to a six year old, you don’t understand it yourself.” I feel that this quote embodies the idea behind this challenge. By using monosyllabic words, the proof condenses down into a simple idea. Hopefully, the proof is just a bit clearer now.<\/p>\n

 <\/p>\n

References:<\/strong><\/p>\n

Stephen Abbot, Understanding Analysis 2nd Edition<\/i><\/p>\n

Tag, By. “A Quote by Albert Einstein.” Goodreads<\/i>. N.p., n.d. Web. 28 Oct. 2016.<\/p>\n","protected":false},"excerpt":{"rendered":"

The Challenge:\u00a0Write a monosyllabic version of a proof we have studied this semester. Accepting the Challenge:\u00a0My strategy was to choose a proof that was visually easy to understand and hope that I can convey the basic idea of the proof to the reader as an image. The proof I chose was the Nested Interval Property. […]<\/p>\n","protected":false},"author":3024,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":true,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2}},"categories":[44423],"tags":[],"class_list":["post-515","post","type-post","status-publish","format-standard","hentry","category-fall-2016"],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p7L4E1-8j","jetpack-related-posts":[],"_links":{"self":[{"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/posts\/515","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/users\/3024"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/comments?post=515"}],"version-history":[{"count":0,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/posts\/515\/revisions"}],"wp:attachment":[{"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/media?parent=515"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/categories?post=515"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/tags?post=515"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}