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So let $(X,d)$ be a metric space. I have to either prove or disprove the following statements. Each of them separately. 1) For any bounded subset $A\subseteq X$ this is true: $\mathrm{diam}(A)=\mathrm{diam}( \overline{\rm A})$

2) For any bounded subset $A\subseteq X$ with non-empty interior it's true that $\mathrm{diam}(A)=\mathrm{diam}(\mathrm{int}(A))$

So I know what interior is: $\mathrm{Int}(A)= \{x\in M\;\;|\;\; \exists r > 0 \mathrm{\;such\;that\;} B_{r}(x) \subset A \}$

So I have to use this for the second one .

So radically, I don't know how to start to solve this, any hints even would help, but I woupd really need help with it.

Thank you in advance.

ts375_zk26
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MathIsTheWayOfLife
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  • Try writting down the definition of the diameter of a set. That might give you some ideas. For instance, it is clear that $\operatorname{diam}(A)\leq \operatorname{diam}(\overline{A})$, do you see why? – Darío G Apr 10 '17 at 09:34

2 Answers2

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1) If you have a point of $\bar{A}$, how did it get there? Recall the definition of a closure, and the definition of diameter. Try to prove that, indeed, the diameter of a closure is the same as the diameter of the set.

2) Think about a set made up of some open ball and another point, far away.

Alon Amit
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(1) is TRUE.

Hint: Let $a,b\in \bar{A}$ be two points. Then there exist sequences $\{a_n\}\subseteq A$ and $\{b_n\}\subseteq A$ such that $a_n\rightarrow a$ and $b_n\rightarrow b$. Then $$d(a,b)\leq d(a,a_n)+d(a_n,b_n)+d(b,b_n)$$ As $a_n\rightarrow a$ and $b_n\rightarrow b$, we have $d(a_n,b_n)\rightarrow d(a,b)$.

(2) is FALSE.

Consider $A=(\mathbb{Q}\cap [0,1])\cup(0,1/2)$.

learning_math
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