# Basic Topology: “A” an open set, then B= A \{p} is an open set

Is this demonstration right or am I missing something?

A an open set, p ∈ A, then B=A\{p} is an open set

Dem:

• x ∈ B
• $\varepsilon_{max} > 0 : B(x,\varepsilon_{max}) ⊂ B$
• $\delta$ = min{ d(p,x) , $\varepsilon_{max}$ }

⇒ B(x,$\delta$) ⊂ B , $\forall$ x ∈ B

Picture

That's it, thanks.

• What is the situation here? What is the topology? Are you in a metric space? I think you have all the right ideas but it is hard to know without a bit more detail. – User8128 Aug 9 '16 at 20:01
• Hint: In a metric space, a singleton set is closed – user251257 Aug 9 '16 at 20:02
• The proof is good – Tsemo Aristide Aug 9 '16 at 20:03
• A-{x}= U{B(a,d(a,x)/2)| a is in A-{x}} is open. – Jacob Wakem Aug 9 '16 at 20:07

Note that finite intersections of open sets are open. Note that $A \setminus \{x\}=A \cap \{x\}^c$. Thus $A \setminus \{x\}$ will be open if $\{x\}^c$ is open, i.e if $\{x\}$ is closed. If you're in a metric space, this is certainly the case.
This demonstration is inadequate, since it assumes that topology $A$ is equiped with a metric (after all, you use $d(p,x)$).
Without further knowledge (e.g. that we are in a metric space) this is certainly not true. As a simple example, if $X = \{a, b\}$ then we have the trivial topology $\tau = \{\emptyset, \{a, b\}\}$ where $\{a, b\} \setminus \{a\} = \{b\} \not\in \tau$.
Generally, it is sufficient that singleton sets are closed (see the answer by mb-); this is the case in $\text{T}_1$ spaces. The trivial topology described above is not even a $\text{T}_0$ space; for a $\text{T}_0$ space which works as a counterexample try $\tau = \{\emptyset, \{a\}, \{a, b\}\}$.