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Let $X$ be a normed linear space and let $Y$ be a closed subspace of $X$. Suppose $(y_n)$ be a sequence in $Y$ such that $d(y_n, A)\to 0$ for a subset $A$ of $X$. Is it true that $d(y_n, A\cap Y)\to 0$? Neither I could prove nor could I find a counterexample. Any hint is appreciated.

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No.

Suppose $X=\mathbb{R}^2$ with the Euclidean norm. Let $Y=\mathbb{R}\times\{0\}$ and $A=\{(0,\frac{1}{n}):n\in\mathbb{N}\}$.

Now take $y_n = (\frac{1}{n},0)$ then $d(y_n,A) = 0$ because $y_n$ is arbitrarily close to $(0,0)$ which is arbitrarily closed to an element in $A$. But $A\cap Y=\emptyset$ so $d(y_n,Y\cap A)$ is not even defined. (If you want that $d(y_n,Y\cap A)$ will be defined just add any element that is not $(0,0)$ to $A$).

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  • $\begingroup$ I am sorry I didn't mention that $A\cap Y\neq \emptyset$. $\endgroup$ – Anupam Jun 9 '19 at 19:01
  • $\begingroup$ @Anupam Just add $(2,0)$ to $A$ and that's it. $\endgroup$ – Mark Jun 9 '19 at 19:03
  • $\begingroup$ Yes, thanks. I got it. $\endgroup$ – Anupam Jun 9 '19 at 19:18

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