Prove that the complement of a point in a metric space is open.

Question

Caveat Lector : This has been done before, but I don’t slavishly want to copy.

Let $A=M-\lbrace x\rbrace$ where $M$ is a metric space. Let $A$ be complementary point and $A \subset M$. $\lbrace

Kaplansky also mentions the closed ball without really defining it.

I have to show Ais open.

Here is 1 approach: Let the given open ball be $B_r(x)=\lbrace D(x,y) < r\rbrace$. If $y\in A$ then $y\in M$ then $B_r(x)$ is contained in $A$ (can we say subset?) so $M$ is open.

2nd approach: To show A is open Show $A^{c}$ is inside a closed ball. This is doable to me

It’s been edited and changed. I hope it’s better. I have trouble using it with curly braces and things — , Feb 06 '20 at 20:36

Prabath Hewasundara · Answer 1 · 2020-02-06T20:56:41.923

1

If $A = \emptyset$ then we are done. if $A\neq\emptyset$,

Let $a\in A$. Then set $r=\frac{d(a,x)}{2}$. Then $x\notin B_r(a)$. Thus $B_r(a)\subseteq A$. Thus $a$ is an interor point of $A$. Since $a$ is arbitrarily chosen, each point of $A$ is an interor point. Thus $A$ is open.

edited Feb 06 '20 at 20:56

answered Feb 06 '20 at 20:38

Prabath Hewasundara

1,644

I cannot assume interior points , no definition in the text – Feb 06 '20 at 20:45
@larrymintz How did you define an open set? – Prabath Hewasundara Feb 06 '20 at 20:51
M is a metric space, x a point. r. the radius. D(x,y) < r The set y € M is called an open ball (Using Kaplansky’s def.) – Feb 06 '20 at 21:05
That's not the definition of open set... – J. De Ro Feb 06 '20 at 21:06
@larrymintz check your notes again. I'm pretty sure that's not the definition of an open set. – Prabath Hewasundara Feb 06 '20 at 21:09
That’s an open ball. Theorem 27 Any open ball in M is an open set By this is theorem I guess to rely on triangle inequality I assumed if y€. A I could I deduced M is open being a metric space but I guess I have to prove it – Feb 06 '20 at 21:10
@larrymintzYes. Any open ball is an open set. But not necessarily vice versa. What is the text anyway? – Prabath Hewasundara Feb 06 '20 at 21:12
Set theory & metric spaces Kaplansky So y€M then implies M has an open ball and have to prove both parts are open – Feb 06 '20 at 21:14
Page 72: Definition. Let $U$ be a subset of a metric space $M$. We say that $U$ is open in $M$ if for every $x\in U$ an entire open ball $S_r(x)$ is contained in $U$.
That is what I have shown.
– Prabath Hewasundara Feb 06 '20 at 21:20
I suppose the radius can be anything as long as they are not equal – Feb 06 '20 at 23:48
No, you cannot take any radius you want. for instance if you take $r=d(a,x)+1$, then $B_r(a) \nsubseteq A$. – Prabath Hewasundara Feb 07 '20 at 05:48
Let us continue this discussion in chat. – Feb 07 '20 at 12:11
Thanks anyways. Next time I will state hints at best – Feb 07 '20 at 18:33

Prove that the complement of a point in a metric space is open.

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