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I need to show that if {x} is open in a metric space X for all x in X,then all subsets of X are open in X

I am using the definition that a set A is open if ∀a∈A,∃r>0 s.t. Br(a)⊆A. I tried proving this by induction, by adding points to the set and since the arbitrary union of open set is open, every subset of X is open. Is this correct?

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  • $\begingroup$ induction won't suffice here as $X$ might have cardinality larger than that of the integers. but since, as you said, an arbitrary union of open sets is open, you don't need to add the points one by one anyway. $\endgroup$
    – Dániel G.
    Commented Dec 1, 2014 at 16:10
  • $\begingroup$ ok thanks a lot $\endgroup$
    – user120768
    Commented Dec 1, 2014 at 16:12

2 Answers 2

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To show this directly just use the following:

Let $A$ be any set in $X$ and $a \in A$. Your premise states that $\left\lbrace a\right\rbrace$ is open. Hence, there exists an $r > 0$ such that $B_r(a) \subseteq \left\lbrace a\right\rbrace$. Since, $\left\lbrace a\right\rbrace \subseteq A$ we also have $B_r(a) \subseteq A$.

It follows that $A$ is open.

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Take any subset $A$ of $X$

Then $A=\cup_{a\in A}\{a\}$

As each singleton is open and arbitrary union of open sets is open as you said so $A$is open

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