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Proof

(related to Proposition: Properties of the Set of All Neighborhoods of a Point)

• By hypothesis, $(X,\mathcal O)$ is a topological space, $A\in X,$ and $\mathcal N(A)$ is the set of all neighborhoods of $A.$

Ad $(1)$

• $X\in\mathcal N(A)$.
• Therefore, $\mathcal N(A)$ is not empty).

Ad $(2)$

• $A\not\in\emptyset$.
• Therefore, $\emptyset$ is not an element of $\mathcal N(A).$

Ad $(3)$

• Assume, $U,W\in\mathcal N(A).$
• By definition of neighborhoods, $A\in O_U\subset U$ and $A\in O_W\subset W$ with $O_U,O_W\in \mathcal O.$
• Since $X$ is a topological space, $\mathcal O$ contains the intersection $O_U\cap O_W.$
• On the other hand, $A\in O_U\cap O_W\subset U\cap W.$
• It follows that $U\cap W$ is a neighborhood of $A.$
• Thus, $U\cap W\in\mathcal N(A).$

Ad $(4)$

• Assume, $U\in\mathcal N(A).$
• Thus, $A\in O\subset U$ for some $O\in \mathcal O.$
• But then $A\in O\subset U\subset W$ for any superset $W$ of $U$.
• It follows that $W$ is a neighborhood of $A.$
• Thus $W\in\mathcal N(A).$
  ∎

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References

Bibliography

1. Grotemeyer, K.P.: "Topologie", B.I.-Wissenschaftsverlag, 1969