MHB The Supremum of a Set is not an Interior Point

[OhMyMarkov]

Hello everyone!

Given a set A that has a supremum $\alpha$, I want to show that $\alpha \notin int(A)$. Is the following proof accepted?

$\alpha = \sup A$ so $\alpha$ is a limit point of $A$. If $\alpha \notin A$, we are done.

Otherwise, for $\forall r>0$, we have $N(\alpha,r)-\{\alpha\} \cap A \neq \varnothing$. Choose $t\in N(\alpha,r)$, and $\alpha<t<\alpha + r$. But for $\forall x \in A$, we have $x\leq\alpha < t$. So $t\notin A$, $t\in A^c$. Therefore, $N(\alpha,r)\cap A^c \neq \varnothing$, so $N(\alpha,r)\not\subset A \; \; \forall r>0$.

Therefore, $\alpha$ is not an interior point of $A$.Is there anything wrong with the proof. I am concerned about the part in red.

Thanks!

 

[Euge]

The statement is not well-formed. Is $A$ a nonempty subset of the real line and bounded above? Even if this is the case, the argument is not correct. Since you want to argue to contradiction, start by assuming to the contrary that $\alpha \in \operatorname{int}A$. Then $A$ contains $N(\alpha, r)$ for some $r > 0$. Since $\alpha + r/2\in N(\alpha,r)$, then $\alpha + r/2 \in A$ and hence $\alpha + r/2 \le \alpha$, a contradiction.

 

[mathwonk]

@Euge: I think I feel a little more generous to this argument. I.e. although there are errors, I believe they are not too egregious. First, perhaps the OP was thinking that since the empty set has no least upper bound, ( every number is an upper bound), and an unbounded above set has no upper bound at all, that the assumption of existence of a least upper bound (sup) implies the set is non empty and bounded above. he/she might however have stated the set lies in the real numbers.

And it seems to me the argument is essentially correct, since it shows that every nbhd of alpha fails to lie in A, hence alpha is not an interior point. I.e. it has shown (directly) that if alpha is the sup, then it is not interior. whereas you have indeed shown (by contradiction) the contrapositive, that if alpha is an interior point then it is not the sup.

Of course, in the OP's opening argument, it is false that the sup of A is necessarily a limit point of A (e.g. consider a finite set A). Perhaps this bothered you. But fortunately this was not used in the argument. So it seems to me that this argument, although it contains a false statement, does go on to give a correct agument independent of that statement. But it should have jumped directly from "[assume] alpha equals sup A" to "Choose [some] t..."

It is also unnecessary to separate off the case where alpha does not belong to A, since the final argument covers this case as well, but this redundancy is not an error.

The key logical point is indeed, as the OP believed, in the red lettered section, and it hinges on the meaning of the phrase "choose t..." What must be shown is that there is a point t in that interval, which then will be shown not to lie in A. While it is obvious that the interval is non empty, and that one can thus choose such a point, it would be more airtight to specifically say "let t = alpha+r/2." ( I.e. the essential thing is not to show that every point of that interval fails to lie in A, but rather than some point fails to, and the second statement does not follow from the first if the interval were empty. The language used does not distinguish clearly between "for some t" and "for all t". Although perhaps "choose t" suggests "for some t", if that is meant, one should justify the choice by remarking first that the interval is non empty.)

So the question seems to me almost well posed (assuming only A is a subset of the reals), and the OP's argument at least essentially contains a correct one, modulo the obvious fact that any interval of form (a,b) with a < b, is non empty. Your argument is of course flawless. cheers!