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I was wondering what can be said about the interior of $\{{4}\}$, the empty set?

The interior of a set $A$ is the largest open set contained by $A$. Hence, if the set at hand is a singleton, then isn't the interior of the singleton the empty set?

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    $\begingroup$ it sounds like you're talking about $\mathbb{R}$ with the standard topology, right? You should specify. There are definitely contexts where the interior of a singleton is the singleton itself, such as in any discrete space $\endgroup$ Nov 12 '12 at 1:41
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It depends on the topology.

If the singleton $\{4\}$ is not an open set in your topology, then the only open subset of $\{4\}$ is $\emptyset$. Thus, $$ \operatorname{Int}(\{4\}) = \bigcup \{\text{open sets } O \mid O \subseteq \{4\}\} = \bigcup \{\emptyset\} = \emptyset. $$ For example, this is the case in the standard (Euclidean) topology on the reals.

On the other hand, if $\{4\}$ is an open set in your topology, then $$ \operatorname{Int}(\{4\}) = \bigcup \{\text{open sets } O \mid O \subseteq \{4\}\} = \bigcup \{\emptyset, \{4\} \} = \{4\}. $$ For example, this is the case in the discrete topology on the reals.

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It depends on what you mean by "open." Since $\lbrace 4\rbrace $ is open with respect to the topology on $\lbrace 4\ \rbrace$, $\text{int}(\lbrace 4 \rbrace)=\lbrace 4\rbrace$. If, for example, you mean open in $\mathbb{R}$, then $\lbrace 4\rbrace=[4,4]$, which is closed and not open, so $\text{int}(\lbrace 4\rbrace)=\emptyset$.

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Remember that when a set $A$ is open $int(A) = A$. Therefore, it largely depends on the topology that you're dealing with and how it defines open sets. Consider the discrete topology and check how singletons behave.

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