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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

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That's it, thanks.

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    $\begingroup$ 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. $\endgroup$ – User8128 Aug 9 '16 at 20:01
  • $\begingroup$ Hint: In a metric space, a singleton set is closed $\endgroup$ – user251257 Aug 9 '16 at 20:02
  • $\begingroup$ The proof is good $\endgroup$ – Tsemo Aristide Aug 9 '16 at 20:03
  • $\begingroup$ A-{x}= U{B(a,d(a,x)/2)| a is in A-{x}} is open. $\endgroup$ – Jacob Wakem Aug 9 '16 at 20:07
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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.

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This demonstration is inadequate, since it assumes that topology $A$ is equiped with a metric (after all, you use $d(p,x)$).

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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 $\tau = \{\emptyset, \{a\}, \{a, b\}\}$.

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