Let $[a,b]$ be any finite closed interval.

(i) $\text{Int}_{[a,b]}(a,b]$

Am I correct to say that the interior of this set is $[a,b]$? Since the interior of a set are all the points in the set in which we can construct an open ball centered at that point which is completely contained inside the set - then for example an open ball centered at $b$ would be completely contained in $(a,b]$, since with respect to $[a,b]$, no part of the ball exists outside this interval?

(ii) $\overline{[a,b)}_{[a,b)}$

Similarly, is the closure of $[a,b)$ here, $[a,b)$? Since although any open ball centered at $b$ has intersection with $[a,b)$ - the point $b$ does not exist in the space?



(i) The interior of the set $(a,b]$ with respect to $[a,b]$ is $(a, b]$. Perhaps you mistyped the left parenthesis? Your reasoning for why there exists an open ball around $b$ which is full contained within $(a,b]$ is correct. But $a$ cannot be in the interior, since $a$ is not even in $(a,b]$ in the first place.

(ii) The closure of $[a,b)$ with respect to $[a,b)$. Here is another way to think about this (this way, I think, better illuminates the key concept of the importance of the larger metric space).

Recall that the closure of a set $A \subseteq X$ is the set of all the limit points of $A$. Recall that a limit point of $A$ is a point in $X$ (the larger metric space!) that has a sequence $a_n \in A$ converging to it.

Here, $b$ cannot be a limit point (and thus cannot be part of the closure) since $b$ is not even in the larger metric space $[a,b)$ in the first place. $\square$

  • $\begingroup$ Hi, yes that was a typo actually. Thanks. $\endgroup$ – JackReacher Apr 15 '15 at 7:00

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