{"id":231447,"date":"2025-06-11T06:30:25","date_gmt":"2025-06-11T06:30:25","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=231447"},"modified":"2025-06-11T06:30:27","modified_gmt":"2025-06-11T06:30:27","slug":"let-a-be-a-countable-set-and-b-be-an-uncountable-set-i-is","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/11\/let-a-be-a-countable-set-and-b-be-an-uncountable-set-i-is\/","title":{"rendered":"Let A be a countable set and B be an uncountable set. (i) Is"},"content":{"rendered":"\n<p>Let A be a countable set and B be an uncountable set. (i) Is<br>countable or uncountable? Explain your answer. (ii) Is<br>countable or uncountable? Explain your answer. (iii) Is<br>countable or uncountable? Explain your answer. [15 marks]<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Here is the solution presented in a formal, textbook-style format.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Let A be a countable set and B be an uncountable set.<\/p>\n\n\n\n<p><strong>(i) Is&nbsp;<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-preformatted\"><strong><code>A\u222aB<em>A<\/em>\u222a<em>B<\/em><\/code><\/strong><\/pre>\n\n\n\n<p><strong>&nbsp;countable or uncountable?<\/strong><\/p>\n\n\n\n<p>The union&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>A\u222aB<em>A<\/em>\u222a<em>B<\/em><\/code><\/pre>\n\n\n\n<p>&nbsp;is&nbsp;<strong>uncountable<\/strong>.<\/p>\n\n\n\n<p><strong>Explanation:<\/strong><br>The union of two sets,&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>A\u222aB<em>A<\/em>\u222a<em>B<\/em><\/code><\/pre>\n\n\n\n<p>, contains all the elements of both A and B. By definition, the set B is a subset of the union&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>A\u222aB<em>A<\/em>\u222a<em>B<\/em><\/code><\/pre>\n\n\n\n<p>, which can be written as&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>B\u2286A\u222aB<em>B<\/em>\u2286<em>A<\/em>\u222a<em>B<\/em><\/code><\/pre>\n\n\n\n<p>.<\/p>\n\n\n\n<p>A fundamental theorem in set theory states that any set which contains an uncountable subset must itself be uncountable. If&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>A\u222aB<em>A<\/em>\u222a<em>B<\/em><\/code><\/pre>\n\n\n\n<p>&nbsp;were countable, then every subset of it would also have to be countable. This would imply that B is countable, which contradicts the initial premise that B is an uncountable set. Therefore, the assumption that&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>A\u222aB<em>A<\/em>\u222a<em>B<\/em><\/code><\/pre>\n\n\n\n<p>&nbsp;is countable must be false. It follows that the union&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>A\u222aB<em>A<\/em>\u222a<em>B<\/em><\/code><\/pre>\n\n\n\n<p>&nbsp;is uncountable. In essence, adding the elements of a countable set to an uncountable set does not change the uncountability of the resulting set.<\/p>\n\n\n\n<p><strong>(ii) Is&nbsp;<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-preformatted\"><strong><code>A\u2229B<em>A<\/em>\u2229<em>B<\/em><\/code><\/strong><\/pre>\n\n\n\n<p><strong>&nbsp;countable or uncountable?<\/strong><\/p>\n\n\n\n<p>The intersection&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>A\u2229B<em>A<\/em>\u2229<em>B<\/em><\/code><\/pre>\n\n\n\n<p>&nbsp;is&nbsp;<strong>countable<\/strong>.<\/p>\n\n\n\n<p><strong>Explanation:<\/strong><br>The intersection of two sets,&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>A\u2229B<em>A<\/em>\u2229<em>B<\/em><\/code><\/pre>\n\n\n\n<p>, consists of all elements that are common to both A and B. By definition, every element of the intersection must be an element of A. This means that the intersection&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>A\u2229B<em>A<\/em>\u2229<em>B<\/em><\/code><\/pre>\n\n\n\n<p>&nbsp;is a subset of A, written as&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>A\u2229B\u2286A<em>A<\/em>\u2229<em>B<\/em>\u2286<em>A<\/em><\/code><\/pre>\n\n\n\n<p>.<\/p>\n\n\n\n<p>A key property of countable sets is that any subset of a countable set is also countable. Since A is given as a countable set, any of its subsets, including&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>A\u2229B<em>A<\/em>\u2229<em>B<\/em><\/code><\/pre>\n\n\n\n<p>, must be countable. This holds true regardless of the nature of B. The intersection could be empty, finite, or countably infinite, but in all possible cases, it cannot be uncountable because it is bounded by the size of the countable set A.<\/p>\n\n\n\n<p><strong>(iii) Is&nbsp;<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-preformatted\"><strong><code>A\u00d7B<em>A<\/em>\u00d7<em>B<\/em><\/code><\/strong><\/pre>\n\n\n\n<p><strong>&nbsp;countable or uncountable?<\/strong><\/p>\n\n\n\n<p>The Cartesian product&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>A\u00d7B<em>A<\/em>\u00d7<em>B<\/em><\/code><\/pre>\n\n\n\n<p>&nbsp;is&nbsp;<strong>uncountable<\/strong>.<\/p>\n\n\n\n<p><strong>Explanation:<\/strong><br>This conclusion assumes A is non-empty. (If A were the empty set,&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>A\u00d7B<em>A<\/em>\u00d7<em>B<\/em><\/code><\/pre>\n\n\n\n<p>&nbsp;would also be empty and thus countable). Since A is a non-empty countable set, it contains at least one element. Let&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>a0<em>a<\/em>0\u200b<\/code><\/pre>\n\n\n\n<p>&nbsp;be a fixed, arbitrary element in A.<\/p>\n\n\n\n<p>Consider the subset of the Cartesian product defined as&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>S={a0}\u00d7B<em>S<\/em>={<em>a<\/em>0\u200b}\u00d7<em>B<\/em><\/code><\/pre>\n\n\n\n<p>. This set consists of all ordered pairs of the form&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>(a0,b)(<em>a<\/em>0\u200b,<em>b<\/em>)<\/code><\/pre>\n\n\n\n<p>, where&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>b<em>b<\/em><\/code><\/pre>\n\n\n\n<p>&nbsp;is any element of B. A one-to-one correspondence (a bijection) can be established between the set B and the set S by the mapping&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>f(b)=(a0,b)<em>f<\/em>(<em>b<\/em>)=(<em>a<\/em>0\u200b,<em>b<\/em>)<\/code><\/pre>\n\n\n\n<p>. Since there is a bijection between B and S, and B is uncountable, the set S must also be uncountable.<\/p>\n\n\n\n<p>The set S is a subset of the full Cartesian product&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>A\u00d7B<em>A<\/em>\u00d7<em>B<\/em><\/code><\/pre>\n\n\n\n<p>&nbsp;(i.e.,&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>S\u2286A\u00d7B<em>S<\/em>\u2286<em>A<\/em>\u00d7<em>B<\/em><\/code><\/pre>\n\n\n\n<p>). As established in part (i), any set that has an uncountable subset is itself uncountable. Since&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>A\u00d7B<em>A<\/em>\u00d7<em>B<\/em><\/code><\/pre>\n\n\n\n<p>&nbsp;contains the uncountable subset S, the Cartesian product&nbsp;<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>A\u00d7B<em>A<\/em>\u00d7<em>B<\/em><\/code><\/pre>\n\n\n\n<p>&nbsp;must be uncountable.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner4-906.jpeg\" alt=\"\" class=\"wp-image-231448\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Let A be a countable set and B be an uncountable set. (i) Iscountable or uncountable? Explain your answer. (ii) Iscountable or uncountable? Explain your answer. (iii) Iscountable or uncountable? Explain your answer. [15 marks] The Correct Answer and Explanation is: Here is the solution presented in a formal, textbook-style format. Let A be a [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","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":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-231447","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231447","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=231447"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231447\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231447"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231447"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231447"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}