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In Are Singleton sets in $\mathbb{R}$ both closed and open?, the best answer says that

The notion of "open" and "closed" depends on a topology you are considering.

Therefore, for example:

  1. In $\mathbb{R}$, the singleton $\{0\}$ is closed.

Now consider the new topological space $A = \{0\} \cup [1,2]$. As in the first paragraph of wiki, the singleton {0} is an open set in the topological space $A$.

Why? I try to answer this in two ways:

  1. $\exists \ \ r$ such that the circle $B_r(0) \in \{0\}$, since we consider only $A$ now.
  2. Since the complement of the open set is closed, $\{0\} = A-[1,2] = [1,2]^c$ is open.

By 1., or 2. we prove that $\{0\}$ is open in the topological space $A$.

My problem is:

If $A = \{0\} \cup (1,2)$, then $\{0\}$ is open or closed?

From 1., it seems open; however, from 2., it seems closed. (Or the reasoning is not correct? )

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sleeve chen
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1 Answers1

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I assume you mean that your set $A = \{ 0 \} \cup (1,2)$ has the subspace topology of $\mathbb R$. In which case the open sets of $A$ are the intersection of open sets in $\mathbb R$ with $A$.

More precisely, if $\tau$ is the collection of open sets of $\mathbb R$, that is the topology of $\mathbb R$, then a topology of $A$ is given by

$$\tau_A = \{ A \cap U \ : \ U \in \tau \}$$

The pairing $(A, \tau_A)$ is the set together with the topology of the set and formally comprise a topological space.

That being the case, $\{ 0 \}$ is an open set in $(A, \tau_A)$ because $\{ 0 \} \in \tau_A$. For example,

$$A \cap \left(-\frac 12, \frac 12\right) = \{ 0 \}$$

The set $(1,2)$ is another open set, as $(1,2) = A \cap (1,2) \in \tau_A$.

Last we see that since $\{ 0 \}$ is the complement in $A$ of the open set $(1,2)$, the set $\{ 0 \}$ is closed.

Simon S
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