# What are necessary and sufficient conditions for a supremum to be a limit point?

A question in similar spirit has already been asked here, but was wrongly accepted.

Consider the case $A = \{0, 1, 2\} \subset \mathbb{R}$, this has $\sup(A) = 2$, but $2$ is not a limit point of $A$.

Note that the answer to this question could be sensitive to the definition of limit, so I give the definition from (Croom 1989): Let $(X,T)$ be a topological space and $A \subset X$, a point $x \in A$ is called a limit point of A if

• every open set containing $x$ contains a point of $A$ distinct from A.

A sufficient condition would be that the supremum does not lays in $A$, e.g. $A = (0,1)$ has $\sup(A) = 1$ and $1$ is a limit point of $A$. But it is not a necesary condition (consider $[ 0, 1 ]$).

Can we say that the supremum is either a maximum or a limit point?

• The answer to your last question is yes, a supremum that is not a maximum has to be a limit point. May 28, 2018 at 14:21

Let $A$ be a nonempty upper-bounded subset of $ℝ$ and let $a := \sup(A)$. There are two options: $a ∈ A$ or $a ∉ A$. There are different two options: there exists $x < a$ such that $(x, a) ∩ A = ∅$ or for every $x < a$ we have $(x, a) ∩ A ≠ ∅$. Let us consider all combinations:

• $a ∈ A$ and $(x, a) ∩ A = ∅$ for some $x < a$: this means that $a$ is an isolated maximum of $A$, like when $A = \{0, 1\}$.
• $a ∈ A$ and $(x, a) ∩ A ≠ ∅$ for every $x < a$: this means that $a$ is a limit maximum of $A$, like when $A = [0, 1]$.
• $a ∉ A$ and $(x, a) ∩ A = ∅$ for some $x < a$: this cannot happen.
• $a ∉ A$ and $(x, a) ∩ A ≠ ∅$ for every $x < a$: this means that $x$ is a non-attained supremum, like when $A = [0, 1)$.

So we have three types of the situation: $\{0, 1\}$, $[0, 1]$, and $[0, 1)$ (isolated maximum, limit maximum, non-attained supremum). The supremum is limit precisely in the last two cases.

Alternatively, let $X$ be a linearly ordered topological space and let $A ⊆ X$. There are five mutually exclussive possibilities regarding $\sup(A)$:

1. $A$ has no supremum.
2. $\sup(A) ∈ A$ and $\sup(A) ∈ \overline{A ∩ (←, \sup(A))}$: limit maximum.
3. $\sup(A) ∈ A$ and $\sup(A) ∉ \overline{A ∩ (←, \sup(A))}$: isolated maximum.
4. $\sup(A) ∉ A$ and $\sup(A) ∈ \overline{A ∩ (←, \sup(A))}$: non-attained supremum.
5. $\sup(A) ∉ A$ and $\sup(A) ∉ \overline{A ∩ (←, \sup(A))}$: $A = ∅$ and $\sup(A) = \min(X)$ … “empty supremum”.

All five cases are realized in $X = [0, ∞)$: $A$ takes the value $[0, ∞)$, $[0, 1]$, $\{0, 1\}$, $[0, 1)$, and $∅$, respectively.

• Can you explain the meaning of: $\overline{A ∩ (←, \sup(A))}$, in particular the $←$, I have used it for $\mathbb{R}$, but never for a linearly ordered topological space. May 28, 2018 at 15:25
• $(←, a)$ is just a shortcut for $\{x ∈ X: x < a\}$. You cannot use $(-∞, a)$ since $-∞$ may already be a point of $X$, e.g. when $X = [-∞, ∞]$. Because of this, $←$ is sometimes used in the interval notation as marker for no value in in the first place of the $(a, b)$ pair. May 28, 2018 at 15:35