# Properties of Sets

I am currently preparing for an exam in functional analysis and I am working through some exercises. However, I do not have the answers for them. Usually I do not need them, but I am still very unconfident with topological terms and the properties of sets, so I hope you can simply tell me, whether my ansers here are correct:

Question is the following: Determine whether the properties (open, closed, compact, bounded) are true for the following sets using the standard Euclidean topology over $\mathbb{R}$:

$\{\frac 1 n + 1 | n \in \mathbb{N}\}$ My thoughts here: All elements are obviously between 2 and 1 (1 excluded), so the set is bounded. However, my problems are to decided whether it is open or closed: For every element of the set, there is no $\epsilon$-environment so that all elements in this environment are also part of the set. For the complement you can always find such an environment. Thus this set is closed. Since it is closed and bounded, it is also compact. True?

Next doubtful point: $\mathbb{Q}$ This is obviously not bounded. Again, you are not able to find $\epsilon$-environments around points in $\mathbb{Q}$. However, you are also not able to find $\epsilon$-environments around points in the complement $\mathbb{R}$ \ $\mathbb{Q}$. Thus it is both open and closed.

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It seems you are making the mistake of assuming "not open" means "closed". But this is not true. In your second argument, you state that there are no $\epsilon$ environments around points in $\Bbb Q$ contained in $\Bbb Q$. This shows that $\Bbb Q$ is not open. This does not imply that $\Bbb Q$ is closed. The second part of your second argument shows that the complement of $\Bbb Q$ is not open. This implies that $\Bbb Q$ is not closed. A set is open iff its complement is closed. –  David Mitra Feb 18 '12 at 14:23
Ok thank you as well - I actually knew that not open is not equal to closed, but I somehow thought that there is no difference between "not open and not closed" and "open and closed" - so thanks for clearing that up –  Chris Feb 18 '12 at 14:28

Your best bet for $\left \{ 1 + \dfrac{1}{n}\, :\, n \in \mathbb{N} \right \}$ is to try and find a sequence in the subset which converges in the space but not in the subset (that is, it converges to a limit outside the subset). Then there is a limit point of the subset which does not lie in the subset, so the subset does not contain all of its limit points, so it is not closed.

Hint: there is a fairly obvious choice of sequence!

What you did wrong with this is assume that for each point in the complement you can find an $\varepsilon$-environment which does not intersect the set. This is not the case, as following through the above will show you: there is one particular point in the complement of the subset about which every $\varepsilon$-environment intersects the subset $-$ can you find it?

As for $\mathbb{Q}$, you've said that it is both open and closed when what you've actually shown is that it is neither open nor closed.

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Yeah, the sequence is $lim_{n\rightarrow\infty} {1 + \frac 1 n} = 1$ - therefore you can't build an environment around 1 - it'll always intersect with the set. Thank you ;) –  Chris Feb 18 '12 at 14:23

For all $n \in \mathbb{N} \smallsetminus \{ 0 \}$ the element $1 + \frac{1}{n}$ is in your set. Is the limit of this sequence in your set? What do you know about limit points and closed sets?

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I know that the limit point is part of the closure of the set, but not the interior. However, the interior in this case is the empty set. So I am not sure what I can conclude from that –  Chris Feb 18 '12 at 14:12
@Chris: A subset $S$ of a topological space $X$ is closed if and only if, for any sequence $(x_n)$ with $x_n \in S$ and $x_n \to x \in X$, we also have $x \in S$. (That is, any sequence in the subset which converges in the space converges in the subset.) Pick any convergent sequence in your subset: is it possible that it can converge to a limit outside the subset? –  Clive Newstead Feb 18 '12 at 14:15