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Considering metric topology and giving the set E subspace topology in the Euclidean Space. Given E={(a$_{1}$, a$_{2}$,...,a$_{n-1}$, 0) | a$_{i}\in\mathbb{R}$} $\subset$$\mathbb{R}^{n}$. I want to determine the limit and interior points of the set E.

Considering the case n=2, E is just the horizontal component $\mathbb{R}$ and since every point is a limit point as well as interior point for $\mathbb{R}$ I feel like the whole set is both interior and the limit points is E $\cup$ {$\infty$}

Any comments would be appreciated.

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  • $\begingroup$ The interior of $E$ in $R^n$ is empty. $\endgroup$ Commented May 4, 2016 at 5:04
  • $\begingroup$ In subspace topology, however, it is the whole A right? Because the open sets in subspace topology don't really go outside E right? $\endgroup$ Commented May 4, 2016 at 5:08
  • $\begingroup$ Any space is open and closed in itself. Any point in E is a limit of a sequence of other points in E. $\endgroup$ Commented May 4, 2016 at 5:10

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You're almost there, but you need to be careful. Look back at your example when $n =2$ and think about the definition of interior point. A point $(x,0) \in E$ is an interior point if there is a small ball $B \subset \mathbb R^2,$ centered at $(x,0)$ with the property that $B \subset E$. But this is impossible. Also note that $\infty$ is not an element of $\mathbb R,$ so that you shouldn't regard this symbol as a limit point. A similar argument holds for the general case.

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  • $\begingroup$ oh because any open set containing point of E basically pops out of E into $\mathbb{R}^{n}$ - E. Right? $\endgroup$ Commented May 4, 2016 at 5:14
  • $\begingroup$ Right. No disk in $\mathbb R^2$ can be a subset of the $x$-axis. $\endgroup$
    – treble
    Commented May 4, 2016 at 5:19

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