Let $S = \{5 - \frac{1}{n} | n \in \mathbb{N} \} \cup (5, \infty) \subset \mathbb{R}$. Determine if $S$ is open, closed or none of that, and if it is compact.
My approach: $S$ is neither open or closed since for example $B_{\epsilon} (4)$ has non-empty intersection with both $S$ and its complement, and similarly for $5$.
Because of the above, $S$ is not compact. Is this correct? Is there anything else needed?
$S$ is not open :$$B_{\varepsilon}(4)\nsubseteq S \text{ for every } \varepsilon$$ and not closed, since the complement is not open:$$B_{\varepsilon}(5)\nsubseteq \mathbb{R}\backslash S \text{ for every } \varepsilon .$$ Since any compact set is closed $S$ is not compact. This is essentially what You pointed out in Your question, so Your argument is entirely sufficient One might also mention that $S$ is not even precompact since it is not bounded and thus not totally bounded.
The boundary of S is equal to {5-1/n : n is element of N} U {5}. Boundary of S is not subset of S. So, S is not closed.
Let us write Int S
Int S = (5,infinity)
Int S is not equal to S. So, S is not open.
I have not learned something about compactness yet in the class .
The expression for $S$ you gave in the second line is wrong. But your reasoning is good.
One knows immediately that $S$ is not closed, because $5$ is an accumulation point and is not in $S$. So $S$ not closed and hence also not compact (Heine- Borel).
For checking if it is open, you can take an element $x$ that satisfies $x=5-\frac 1 m$ for some $m\in\mathbb{N} $. Take a ball $B_k$ around that point with radius $\frac{1}{k}>0$ and verify that there exist some $k_0$ such that for all $k\geq k_0$ we have $x$ is the only element in $S$ that is in $B_k$. Then you can conclude that you cannot find a ball with middle point $x$ such that the ball is contained in $S$. So $S$ not open.
