- #1

OhMyMarkov

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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!

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!

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