{"id":9519,"date":"2022-01-15T23:34:56","date_gmt":"2022-01-15T18:04:56","guid":{"rendered":"https:\/\/mathemerize.com\/?p=9519"},"modified":"2022-01-15T23:34:59","modified_gmt":"2022-01-15T18:04:59","slug":"prove-that-empty-set-is-a-subset-of-every-set","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/prove-that-empty-set-is-a-subset-of-every-set\/","title":{"rendered":"Prove that Empty Set is a Subset of Every Set."},"content":{"rendered":"
Let A be any set and \\(\\phi\\) be the empty set.<\/p>\n
In order to show that \\(\\phi\\) \\(\\subseteq\\) A, we must show that every element of \\(\\phi\\) is an element of A also. But, \\(\\phi\\) contains no element.<\/p>\n
So, every element of \\(\\phi\\) is in A.<\/p>\n
Hence, \\(\\phi\\) is the subset of A.<\/p>\n","protected":false},"excerpt":{"rendered":"
Solution : Let A be any set and \\(\\phi\\) be the empty set. In order to show that \\(\\phi\\) \\(\\subseteq\\) A, we must show that every element of \\(\\phi\\) is an element of A also. But, \\(\\phi\\) contains no element. So, every element of \\(\\phi\\) is in A. Hence, \\(\\phi\\) is the subset of A.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[43,56],"tags":[],"yoast_head":"\n