We were taught that $[a,a]=\{a\}$ and $(a,a)=\emptyset $, For $a \in \Bbb R$.
So I wonder what will be the result in the case $(a,a]$?
Let $A=(a,a]$.
Then $a \in A$ and $a \notin A$. This is absurd. So I guess $A$ is undefined.
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1I think your reasoning is correct. – Max Payne Nov 05 '16 at 07:04
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3Alternately you could view it as the set of points $x$ such that $a<x\leq a$ which would also give $A=\emptyset$. – Ian Miller Nov 05 '16 at 07:07
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@MathematicsStudent1122 but how to implement that idea? Since we get absurdity using it. – Error 404 Nov 05 '16 at 07:13
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1@VikrantDesai That was a mistake. Your idea is actually wrong. It's not true that $a \in A$ because then we'd have $a<a$. The set is just empty. – MathematicsStudent1122 Nov 05 '16 at 07:16
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@MathematicsStudent1122 I think I understood it now. – Error 404 Nov 05 '16 at 07:21
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@MathematicsStudent1122 I don't think the idea is 'wrong'. He wrote the two contradictory conditions that should be satisfied at the same time, which led to absurdity. – Max Payne Nov 05 '16 at 07:28
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2@AsafKaragila None of the answers given by you in the duplicate question specifically address the case of $a=b$. Please reopen this question or update your answer to the linked question to explicitly address the case of $a=b$ and $(a,b]$. – Ian Miller Nov 05 '16 at 07:50
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@MaxPayne please see my comment below Zachery Selk's answer. :) – Error 404 Nov 05 '16 at 07:53
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@Ian: Last I recall, $a\geq b$ includes the case of $a=b$. I might be wrong, though. – Asaf Karagila Nov 05 '16 at 07:53
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@AsafKaragila I got my query clarified. :) But still if I want to re-open the question can I add the tag 'proof-verification' ? – Error 404 Nov 05 '16 at 08:09
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@AsafKaragila It is covered by that case but it is not mentioned explicitly. You only explicitly mention $[a,b]$ for the case $a=b$. – Ian Miller Nov 05 '16 at 08:23
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@Ian: There is an answer to the question in the link. If someone needs clarification that $[a,b)$ or $(a,b]$ are empty when $a\geq b$ includes the case that $a=b$, then that someone doesn't deserve the attention or energy of any community member here. And indeed, it seems that the OP can make that simple inference on their own. – Asaf Karagila Nov 05 '16 at 08:25
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Searching for $(a,a]$ using SE suggest that this would actually be a better candidate for redirection: http://math.stackexchange.com/questions/1384184/is-a-a-emptyset – Ian Miller Nov 05 '16 at 08:25
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@Vikrant: You could add proof-verification, which makes this a slightly different question, to which none of the three answers below is actually relevant. – Asaf Karagila Nov 05 '16 at 08:26
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@Ian: There are probably half a dozen better candidates. I picked the one I had the easiest time to find. Feel free to flag this for a moderator to re-close as a better duplicate. – Asaf Karagila Nov 05 '16 at 08:27
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@AsafKaragila okay no problem. :) – Error 404 Nov 05 '16 at 08:33
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@MaxPayne The assumption $a\in A$ leads to absurdity. That does not mean that $A$ is undefined, but that the assumption $a\in A$ is wrong. Conclusion: $a\notin A$ for every $a$, or equivalently $A=\varnothing$. – drhab Nov 05 '16 at 08:46
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According to the (very reasonable) definition given on Wikipedia:
$$(a,a]:=\{x\in \Bbb{R}:a<x\le a\}=\emptyset$$
I think this is a fair definition.
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So if $x \in A$ then $x \gt a$. Hence $x \neq a$. Also $x \le a$ means $x \lt a$ or $x = a$. But since $x \neq a$ we have $x \lt a$. Therefore we have got $x \lt a$ and $x \gt a$ which renders $A$ empty. Am I correct? – Error 404 Nov 05 '16 at 07:18
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1@VikrantDesai You seem to over complicate it. The set is empty simply because there is no number that is greater than $a$ and at the same time less than or equal to $a$, these two relations are mutual exclusive. – Sil Nov 05 '16 at 09:21
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$A =(a,a]$ is the empty set
Fred
Fred
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Can you provide a reference or justification or something? Otherwise this looks very much like an opinion. – Ian Miller Nov 05 '16 at 07:40
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