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The questions are the following:

Consider the five topologies on the real line $\mathbb R$:

  1. $\mathcal T_1$: the standard topology

  2. $\mathcal T_2$: the $K$-topology

  3. $\mathcal T_3$: the finite complement topology

  4. $\mathcal T_4$: the upper limit topology

  5. $\mathcal T_5$: the topology generated by the basis $\{(-\infty,a)\mid a \in \mathbb R\}$

Determine the closure of the set $K=\{\frac{1}{n}\mid n\in\mathbb N\}$ under each of these topologies.

My answer is the following:

$\mathcal T_1$: $\mathrm{cl}(K)=\{0\} \cup K$.

$\mathcal T_4$: $\{0\} \cup K$.

$\mathcal T_5$: $[0, \infty)$

Thank you.

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9  
Which can you solve? What have you tried for the others, and what problems have you met? –  Chris Eagle Oct 23 '11 at 17:57
7  
math.stackexchange.com is not a black-box. Where do you get stuck? Can you see when a set is closed in the different topologies? –  AD. Oct 23 '11 at 18:18
2  
What they said. The purpose of this problem is to help you learn the definitions. So try to do them yourself. If you want, write your solutions here for comments. –  GEdgar Oct 23 '11 at 18:34
4  
You should also explain what the $K$-topology is. (The others are standard.) –  Brian M. Scott Oct 23 '11 at 20:35
    
Note that in the $K$-topology your set by definition is closed, so what is its closure? Also, in the finite complement topology on an infinite set, every infinite set has the whole space as its closure. This is pretty easy as an exercise: every non-empty open set can only miss finitely many points, so cannot miss the infinite set. –  Henno Brandsma Nov 23 '11 at 10:20

2 Answers 2

$\mathcal{T}_1:$ That's right, well done. Note that this topology is generated by the intervals $(a,b)$. So for any point $x$ in $\mathbb{R}$ in between $k = \frac1n$ and $k^\prime = \frac1m$ we can choose $(x - m, x + m)$ where $m = \frac12 \min (\frac1n , \frac1m)$ to see that $x$ is not in the closure of $K$. Looking at $0$ we see that every interval $0 \in (a,b)$ has to have $0 < b$ so there will be an $n$ such that $0 < \frac1n < b$ and hence every open set containing $0$ will have non-empty intersection with $K$. Hence we get $\overline{K} = K \cup \{0\}$.

$\mathcal{T}_2$: Note that $\mathcal{T}_2$ is generated by $(a,b) $ and $ (a,b) \setminus K $ so $\mathcal{T}_1 \subset \mathcal{T}_2$ and hence $ \overline{K}_{\mathcal{T}_2} \subset \overline{K}_{\mathcal{T}_1}$ since we have more sets that might contain our point but might not intersect with $K$. Since $K \subset \overline{K}$ we therefore know that the only point we need to think about is $0$. But none of the sets we added to the basis intersect with $K$ hence there now are open sets $O$ containing $0$ with $O \cap K = \varnothing$. Hence $0$ is not in the closure of $K$ and we have $\overline{K}=K$.

$\mathcal{T}_3:$ Note that the open sets in this topology look like this: $O = \mathbb{R} \setminus \{x_1, \dots, x_n \}$. Consider a point $p$ in $\mathbb{R}$. Let $O$ be an open set containing $p$. Note that $O$ has non-empty intersection with $K$ since $O$ contains all points of $\mathbb{R}$ except finitely many. In particular, $K$ contains infinitely many points hence their intersection will be non-empty. Since $p$ was arbitrary we get $\overline{K} = \mathbb{R}$.

$\mathcal{T}_4$: Note that the upper limit topology is generated by intervals of the form $(a, b]$. Note that for a point $p \neq 0$ outside $K$ we can pick $(\varepsilon , p]$ with $\varepsilon$ small enough such that $(\varepsilon , p] \cap K = \varnothing$. For $0$ we pick $(a, 0]$ to get a set that does not intersect with $K$ hence $0$ is not in the closure of $K$. Hence we get $\overline{K} = K$.

$\mathcal{T}_5$: Again you have it right, well done. To see this pick a point $p$ to the right of $0$ or $0$ itself and note that every open interval $(-\infty, a)$ containing $p$ will intersect with $K$ since there will be an $n$ such that $\frac1n < p$.

Hope this helps.

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Matt N, $\mathcal T _2$ is wrong. $0$ is not in the closure of $K$, once $K$ is not in any neighborhood $(-\epsilon, \epsilon) - K$ of $0$. –  user73051 Apr 17 '13 at 14:29
    
Indeed. Thank you! I will correct it right away. For the future: you should use the comment function instead of "answer" when you have a comment to a post. –  Matt N. Apr 17 '13 at 14:39
    
@stsz: Welcome to MSE! I realize you don't yet have enough reputation, but this is better left as a comment as opposed to an answer. Regards –  Amzoti Apr 17 '13 at 14:52

You just use the definition of the closure, for example: $x \in cl(K)$ if and only if every open neighborhood $V$ containing $x$ intersects $K$. Use the obvious $K \subset cl(K)$. Then think of any possible limit points by looking at the form of the neighborhoods.

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