Let $A$ be a subset. If $A$ is non-empty, then $A$ has at least two subsets. Specifically, $emptyset$...












0












$begingroup$


Still new to proofs so walking through this step-by-step.



Claim: If $A$ is a non-empty, then $A$ has at least two subsets. Specifically, $emptyset$ $subseteq A$ and $A subseteq A$.



Proving the contrapositive.



If $A$ has less than two subsets, then $A$ is empty.



Would that just imply that $emptyset subseteq A$, which means that the empty set is the only set in $A$?



I know I haven't provided much, but I just need a push in the right direction. Still brand new to elementary set-theory.










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    No, just argue directly.
    $endgroup$
    – Andrés E. Caicedo
    Jan 7 at 21:25






  • 3




    $begingroup$
    We always have $emptyset subseteq A$ and $A subseteq A$. If $A$ has only 1 subset, then $A = emptyset$.
    $endgroup$
    – Paul K
    Jan 7 at 21:26
















0












$begingroup$


Still new to proofs so walking through this step-by-step.



Claim: If $A$ is a non-empty, then $A$ has at least two subsets. Specifically, $emptyset$ $subseteq A$ and $A subseteq A$.



Proving the contrapositive.



If $A$ has less than two subsets, then $A$ is empty.



Would that just imply that $emptyset subseteq A$, which means that the empty set is the only set in $A$?



I know I haven't provided much, but I just need a push in the right direction. Still brand new to elementary set-theory.










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    No, just argue directly.
    $endgroup$
    – Andrés E. Caicedo
    Jan 7 at 21:25






  • 3




    $begingroup$
    We always have $emptyset subseteq A$ and $A subseteq A$. If $A$ has only 1 subset, then $A = emptyset$.
    $endgroup$
    – Paul K
    Jan 7 at 21:26














0












0








0





$begingroup$


Still new to proofs so walking through this step-by-step.



Claim: If $A$ is a non-empty, then $A$ has at least two subsets. Specifically, $emptyset$ $subseteq A$ and $A subseteq A$.



Proving the contrapositive.



If $A$ has less than two subsets, then $A$ is empty.



Would that just imply that $emptyset subseteq A$, which means that the empty set is the only set in $A$?



I know I haven't provided much, but I just need a push in the right direction. Still brand new to elementary set-theory.










share|cite|improve this question









$endgroup$




Still new to proofs so walking through this step-by-step.



Claim: If $A$ is a non-empty, then $A$ has at least two subsets. Specifically, $emptyset$ $subseteq A$ and $A subseteq A$.



Proving the contrapositive.



If $A$ has less than two subsets, then $A$ is empty.



Would that just imply that $emptyset subseteq A$, which means that the empty set is the only set in $A$?



I know I haven't provided much, but I just need a push in the right direction. Still brand new to elementary set-theory.







elementary-set-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 7 at 21:24









RyanRyan

1356




1356








  • 1




    $begingroup$
    No, just argue directly.
    $endgroup$
    – Andrés E. Caicedo
    Jan 7 at 21:25






  • 3




    $begingroup$
    We always have $emptyset subseteq A$ and $A subseteq A$. If $A$ has only 1 subset, then $A = emptyset$.
    $endgroup$
    – Paul K
    Jan 7 at 21:26














  • 1




    $begingroup$
    No, just argue directly.
    $endgroup$
    – Andrés E. Caicedo
    Jan 7 at 21:25






  • 3




    $begingroup$
    We always have $emptyset subseteq A$ and $A subseteq A$. If $A$ has only 1 subset, then $A = emptyset$.
    $endgroup$
    – Paul K
    Jan 7 at 21:26








1




1




$begingroup$
No, just argue directly.
$endgroup$
– Andrés E. Caicedo
Jan 7 at 21:25




$begingroup$
No, just argue directly.
$endgroup$
– Andrés E. Caicedo
Jan 7 at 21:25




3




3




$begingroup$
We always have $emptyset subseteq A$ and $A subseteq A$. If $A$ has only 1 subset, then $A = emptyset$.
$endgroup$
– Paul K
Jan 7 at 21:26




$begingroup$
We always have $emptyset subseteq A$ and $A subseteq A$. If $A$ has only 1 subset, then $A = emptyset$.
$endgroup$
– Paul K
Jan 7 at 21:26










1 Answer
1






active

oldest

votes


















1












$begingroup$

If A is empty, then A does not have two subsets.



Thus, if A has two subsets, A is not empty.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is the converse of what the OP is trying to prove.
    $endgroup$
    – Barry Cipra
    Jan 8 at 12:30











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065514%2flet-a-be-a-subset-if-a-is-non-empty-then-a-has-at-least-two-subsets-spe%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

If A is empty, then A does not have two subsets.



Thus, if A has two subsets, A is not empty.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is the converse of what the OP is trying to prove.
    $endgroup$
    – Barry Cipra
    Jan 8 at 12:30
















1












$begingroup$

If A is empty, then A does not have two subsets.



Thus, if A has two subsets, A is not empty.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is the converse of what the OP is trying to prove.
    $endgroup$
    – Barry Cipra
    Jan 8 at 12:30














1












1








1





$begingroup$

If A is empty, then A does not have two subsets.



Thus, if A has two subsets, A is not empty.






share|cite|improve this answer











$endgroup$



If A is empty, then A does not have two subsets.



Thus, if A has two subsets, A is not empty.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 8 at 9:35









Math_QED

7,52831450




7,52831450










answered Jan 8 at 9:31









William ElliotWilliam Elliot

7,8502720




7,8502720












  • $begingroup$
    This is the converse of what the OP is trying to prove.
    $endgroup$
    – Barry Cipra
    Jan 8 at 12:30


















  • $begingroup$
    This is the converse of what the OP is trying to prove.
    $endgroup$
    – Barry Cipra
    Jan 8 at 12:30
















$begingroup$
This is the converse of what the OP is trying to prove.
$endgroup$
– Barry Cipra
Jan 8 at 12:30




$begingroup$
This is the converse of what the OP is trying to prove.
$endgroup$
– Barry Cipra
Jan 8 at 12:30


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065514%2flet-a-be-a-subset-if-a-is-non-empty-then-a-has-at-least-two-subsets-spe%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

'app-layout' is not a known element: how to share Component with different Modules

android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

WPF add header to Image with URL pettitions [duplicate]