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Let $x_n \to x$ in Metric Space $(X,d)$ and A is a subset of X. Show that $d(x_n,A) \to d(x,A)$.

I have used the generalized triangle inequality using an element $a$ of A. How do I show that the infimums of over all $a$ belonging to A would satisfy the condition and ultimately,$d(x_n,A) \to d(x,A)$?

Infinity_hunter
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Sepia Glen
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You can just show that the function $f_A(x)=d(x,A):X\longrightarrow\mathbb{R}_+$ is continuous (clearly $A$ is assumed non-empty). This follows since $|f_A(x)-f_A(y)|\leq d(x,y)$ (mathematicians call such function Lipschitz continuous with constant 1, just so you know). Indeed for any $a\in A$, $d(x,a)\leq d(x,y)+d(y,a)$ and so, taking the $\inf_A$ gives $f_A(x)\leq d(x,y)+d(y,a)$. Rearrange to $d(y,a)\geq f_A(x)-d(x,y)$. Taking $\inf_A$ again yields $f_A(y)\geq f_A(x)-d(x,y)$ and so $f_A(x)-f_A(y)\leq d(x,y)$. There is nothing special about the order of $x,y$ as $d(x,y)=d(y,x)$ so reversing their roles in the previous conclusion gives also $f_A(y)-f_A(x)\leq d(x,y)$. Hence $|f_A(x)-f_A(y)|\leq d(x,y)$. This means $f_A$ is continuous, as you can just take $\delta=\epsilon$ in the definition of continuity. The continuity of $f_A$ directly implies your claim (by the sequential definition of continuity, if you want).

xyz
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  • I do understand your way but I haven't learnt about the Lipschitz continuity. So, is there any other way to solve it? – Sepia Glen Sep 20 '21 at 12:04
  • You can ignore the term for now, it's just the name mathematicians give to the inequality $|f_A(x)-f_A(y)|\leq d(x,y)$. As long as you understand how that inequality is derived, it is all good, whether you know about Lipschitzianity of functions or not. It is just a name. – xyz Sep 20 '21 at 12:15
  • I added a couple of details to bypass the mention of Lipschitzianity, which is now in the brackets, so you feel more comfortable about it. – xyz Sep 20 '21 at 12:20
  • Thanks a lot! It makes sense to me now. – Sepia Glen Sep 21 '21 at 09:14