Let $X=\mathbb{R}$ . Let $$T = \{X,\ \emptyset,\ \text{ all subsets of $X$ of the form }(a,\infty) = \{x \in X|a<x\}\text{ where }a\in \mathbb{R}\}$$ Is $T$ a topology on $X$?

What is the closure, interior, and boundary of $[3,7)=\{x\mid 3 \leq x < 7\}$?

Same question for $\{1,3,5\}$.

Let $A$ $=$ $\{1,3,5\}$ I know that $A$ is a closed subset of $x$. Therefore it can be itself. The interior is the $\emptyset$ and the boundary must be $A$. This I think I got it.

Here is what I know:

This is a topology. The closure of $[3,7)$ is the set $(-\infty ,7]$. The closure of $\{1,3,5\}$ is $(-\infty,5]$.

Not sure about the interior and boundary. Can someone please help me to see this?

  • $\begingroup$ Just editted it $\endgroup$ – user99744 Nov 12 '13 at 18:40
  • $\begingroup$ Under what topology? $\endgroup$ – LASV Nov 12 '13 at 18:41
  • $\begingroup$ a is any real number. x is the set in the topology where $x$ belong to $T$ $\endgroup$ – user99744 Nov 12 '13 at 18:44
  • $\begingroup$ The interior is defined to be the union of all the open sets contained in $[3,7)$. Are there any? $\endgroup$ – LASV Nov 12 '13 at 18:46
  • $\begingroup$ @Luis. There is no open set since it is not an open interval $\endgroup$ – user99744 Nov 12 '13 at 18:49

The interior of $[3,7)$ is empty, since each non-empty open set is unbounded. The closure of this set is $(-\infty,7]$, as each point $x\le7$ has only neighborhoods $(a,∞)$ with $-∞\le a<7$. Since the boundary is the closure minus the interior, it must be $(-∞,7]$ again.

The set $B=\{1,3,5\}$ has empty interior, and its closure is $(-∞,5]$.

  • $\begingroup$ Thanks! That is what I was looking for. I knew I was somewhat on the correct path but could not figure it out. Thanks again. $\endgroup$ – user99744 Nov 12 '13 at 20:15

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