Prove: $A\subseteq \mathbb{R}$ is an upper bounded set that contains at least two items/numbers. If $x < \sup{A}$, then $\sup{\left(A\setminus\{x\}\right)}=\sup A$.

My attempt:

Since $A$ is upper bounded, and non-empty, $A$ has a supremum according to the upper bound property, which we'll denote with $\sup{A}=c$. Since $c$ is $\sup{A}$, for all $a\in A$, $a< c$.

Therefore $\sup{A\setminus\{x\}}=\sup{A}$.

My questions is: Is my though process correct? I feel like I may not be understanding the question correctly, and this seemed to be too easy for an exam question.

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    $\begingroup$ Do you mean "upper bounded set" not "upper bounded group" because the notion of group isn't needed in this exercise. $\endgroup$ – Scientifica Jun 25 '16 at 13:20
  • $\begingroup$ It appears the direct translation is "set" $\endgroup$ – RonaldB Jun 25 '16 at 13:21
  • $\begingroup$ I didn't understand but these will help you to see the difference: en.wikipedia.org/wiki/Set_(mathematics) en.wikipedia.org/wiki/Group_(mathematics) $\endgroup$ – Scientifica Jun 25 '16 at 13:23
  • $\begingroup$ The word used in my language for group (non-mathematical) and set (mathematical) is the word group. What I meant to say is that it should be set. $\endgroup$ – RonaldB Jun 25 '16 at 13:24
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    $\begingroup$ Thank you for your answer. I don't speak Hebrew sorry, but in general, in a language you will find many synonyms for the same word. In my opinion, the word in Hebrew that means "set" should be the one that simply refers to a collection of things, while the word for "group" should be a synonym from which we infer there's a certain relation between the elements. Nevertheless, according to Wikipedia, there are two different words used for the mathematical words group and set in Hebrew: קבוצה for set and חבורה for group. $\endgroup$ – Scientifica Jun 25 '16 at 13:42

Firstly, you made a little mistake in

Since $c$ is $\sup{A}$, for all $a\in A,\,a<c$

It should be

Since $c$ is $\sup{A}$, for all $a\in A,\,a\le c$

To convince you, simply take the case where $\sup{A}\in A$ like in the case where $A=[1,2]$. Here $\sup{A}=2\in A$ and $\forall a\in A,\,a\le 2$.

Secondly, your conclusion

Therefore $\sup{A\setminus\{x\}}=\sup{A}$

seems quite unclear to me. You didn't explain what you did with $c'=\sup{A\setminus\{x\}}$.

Thirdly, when you want to prove the equality of the $\sup$ of two different sets (here $c$ for one set as you named it and $c'$ for the other one as I named it) , I recommend you to, in general, try to prove that $c\le c'$ and that $c'\le c$. $c'\le c$ should be easy since $\left(A\setminus\{x\}\right)\subset A$. To prove $c\le c'$, you must clearly show how you used the assumption $x<\sup{A}$.


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