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I am having some trouble solving the following question in Willard's General Topology (p.36).

At each point $x$ of the real line other than the origin, the basic neighbourhoods of $x$ will be the usual open intervals centred at $x$. Basic neighbourhoods of the origin will be the sets: $(-\epsilon, \epsilon) \cup (-\infty, -n) \cup (n, \infty)$, for all possible choices $\epsilon > 0$ and $n \in \mathbb{N}$.

Q1. Verify that this gives a topology on the line.

Q2. Describe the closure operation in the resulting space.

For Q1, to show that the basic neighbourhoods give a topology on $\mathbb{R}$, I have to show that the following properties hold (from Theorem 4.5 in the textbook):

Edit: as pointed out in the comments, my description of the following theorem was inaccurate before. It has been updated - thank you to Brian M. Scott for pointing this out.

a) If $V \in \mathfrak{B}_x$ then $x \in V$.

b) If $V_1, V_2 \in \mathfrak{B}_x$ then $\exists V_3 \in \mathfrak{B}_x : V_3 \subseteq V_1 \cap V_2$.

c) If $V \in \mathfrak{B}_x$ then $\exists V_0 \in \mathfrak{B}_x$ such that $\forall y \in V_0, \exists W \in \mathfrak{B}_y : W \subseteq V$.

This is not hard to show now that I have the theorem copied correctly. However, I am still having problems with Q2.

How does one go about comprehensively describing the closure operation? Based on the above, the open sets are:

  1. The bounded, open sets $U \subseteq \mathbb{R}$ under the standard topology, where $0 \notin U$.
  2. The open sets $U \subseteq \mathbb{R}$ under the standard topology, for which $0 \in U$ and $(-\infty, a) \cup (b,\infty) \subseteq U$ for some $a<0, b>0$.

Obviously, the closed sets are the complements of open sets. But I am having trouble figuring out a concise way to describe the closure operation. Any further insight is appreciated.

2 Answers2

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You’ve stated (c) incorrectly. In fact, it makes no sense as you’ve stated it: you can’t talk about the interior of a set until you have a topology, and the point of the first question is to show that you do have a base for a topology. Once you’ve done that, you can define the topology using (d) of Willard’s Theorem $4.5$. The correct statement of (c) is:

If $V\in\mathfrak{B}_x$, there is some $V_0\in\mathfrak{B}_x$ such that if $y\in V_0$, then there is some $W\in\mathfrak{B}_y$ such that $W\subseteq V$.

Verifying (a) is easy. If $x\ne 0$, members of $\mathfrak{B}_x$ have the form $(x-r,x+r)$ for some $r>0$, and clearly $x\in(x-r,x+r)$. And if $x\ne 0$, every $V\in\mathfrak{B}_x$ contains a subset of the form $(-\epsilon,\epsilon)$ for some $\epsilon>0$, and clearly $0\in(-\epsilon,\epsilon)\subseteq V$.

Verifying (b) is almost as easy. Suppose that $V_1,V_2\in\mathfrak{B}_x$. If $x\ne 0$ there are $r_1,r_2>0$ such that $V_1=(x-r_1,x+r_1)$ and $V_2=(x-r_2,x+r_2)$. Let $r=\min\{r_1,r_2\}$, and let $V_3=(x-r,x+r)$; then $V_3\in\mathfrak{B}_x$, and $V_3\subseteq V_1\cap V_2$. If $x=0$, there are $n_,n_2\in\Bbb Z^+$ and $\epsilon_1,\epsilon_2>0$ such that $V_i=(-\epsilon_i,\epsilon_i)\cup(\leftarrow,-n)\cup(n,\to)$ for $i=1,2$. Let $\epsilon=\min\{\epsilon_1,\epsilon_2\}$ and $n=\max\{n_1,n_2\}$, and let $V_3=(-\epsilon,\epsilon)\cup(\leftarrow,-n)\cup(n,\to)$; then $V_3\in\mathfrak{B}_0$, and $V_3\subseteq V_1\cap V_2$.

And because it will turn out that all of these basic nbhds are actually open in the topology that they generate, it’s also easy to verify (c). Let $V\in\mathfrak{B}_x$, and suppose first that $x\ne 0$, so that $V=(x-r,x+r)$ for some $r>0$. Let $s=\min\{r,|x|\}$, and let $V_0=(x-s,x+s)\in\mathfrak{B}_x$. If $y\in V_0$, let $t=\min\{(x+s)-y,y-(x-s)\}$; then $(y-t,y+t)\subseteq V_0$, and $(y-t,y+t)\in\mathfrak{B}_y$, since $y\ne 0$. (We needed $s$ in order to ensure that $0\notin V_0$.)

Now suppose that $x=0$, so that $V=(-\epsilon,\epsilon)\cup(\leftarrow,-n)\cup(n,\to)$ for some $\epsilon>0$ and $n\in\Bbb Z^+$. I’ll leave it to you verify that we can take $V_0=V$: show that for each $y\in V$ there is a $W\in\mathfrak{B}_y$ such that $W\subseteq V$.

The second question becomes very easy once you realize that this space is homeomorphic to an $\infty$ sign in the plane, e.g., to the union of the circles of radius $1$ with centres at $\langle -1,0\rangle$ and $\langle 1,0\rangle$. The topology is designed to bend the ends of the real line back around to approach $0$, so that $0$ becomes the centre point of the $\infty$.

See if you can finish it off with that much of a clue.

Brian M. Scott
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  • Thanks, I copied the theorem down wrong and my confusion stemmed from the fact that it didn’t make any sense as I wrote it! As for Q2, I haven’t seen homeomorphisms in the textbook yet (they’re in Ch. 7) although I do know that a homeomorphism is a sort of “topological equivalence”. Still, I’m assuming there’s a way to solve the problem only using results from the textbook so far. I think that the open sets are the intervals of the form $(x-c, x+c)$ which don’t contain $0$, as well as the sets that are of the form $(-\epsilon, \epsilon) \cup (-\infty, n) \cup (n, \infty)$. Is this correct? – TuringTester69 Sep 24 '20 at 22:35
  • Sorry, ran out of words. The above should really say that the open sets are the intervals of the forms I wrote, as well as the unions, finite intersections, the empty set and the real line. – TuringTester69 Sep 24 '20 at 22:38
  • @TuringTester69: More explicitly, every bounded open subset of $\Bbb R$ that does not contain $0$ is open in this space. The open sets in this space that do contain $0$ are the open sets $U$ in $\Bbb R$ that contain $0$ and for which there are $a>0$ and $b<0$ such that $(\leftarrow,b)\cup(a,\to)\subseteq U$. You don’t really need to know what a homeomorphism is to make use of the idea that I suggested: just think of bending the ends of the real line around so that $\lim_{x\to\infty}x=0$ and $\lim_{x\to-\infty}x=0$. Open sets that don’t contain $0$ just get ... – Brian M. Scott Sep 24 '20 at 22:41
  • ... bent a bit, and open sets that contain $0$ have to contain open intervals along all four of the ‘arms’ at $0$. – Brian M. Scott Sep 24 '20 at 22:42
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    I’ve been mulling over it a bit and still can’t figure out how to completely describe the closure operation. Could you please provide some more details? – TuringTester69 Sep 27 '20 at 19:48
  • @TuringTester69: Every bounded set just has its usual closure. Every unbounded set has its usual closure together with $0$. – Brian M. Scott Sep 28 '20 at 03:38
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You haven't really addressed what to do when $y$ is also $0$. In such a case, your parenthetical makes the false claim that we can take $W = (-\varepsilon, \varepsilon)$, but this is not an element of $\mathfrak{B}_y = \mathfrak{B}_0$.

Your handling of $x \neq 0$ is fine.

For $x = 0$, I might write...

For $x = 0$, let $V \in \mathfrak{B}_x = \mathfrak{B}_0$. Then, there are $\varepsilon > 0$ and $n > 0$ such that $V = (-\infty, -n) \cup (-\varepsilon, \varepsilon) \cup (n, \infty)$. Notice $V = V^\circ$, so let $y \in V$.

If $y \in (-\infty, -n) \in \mathfrak{B}_y$, we may take $W = (-\infty, -n) \subseteq V$. A similar argument is used for $y \in (n,\infty) \in \mathfrak{B}_y$.

If $y \in (-\varepsilon, \varepsilon)$, either $y = 0$ or $y \neq 0$. If $y = 0$, let $W = V \in \mathfrak{B}_x = \mathfrak{B}_y = \mathfrak{B}_0$. If $y \neq 0$, take $W = (y/2, 2y) \cap (-\varepsilon, \varepsilon) \in \mathfrak{B}_y$. Since $$ -\varepsilon < -|y| < -|y/2| < |y/2| < |y| < \varepsilon \text{,} $$ this intersection is a nonempty open neighborhood of $y$.

Eric Towers
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  • Thanks, you’re right, I see I didn’t explicitly mention the case when $y = 0$. I’m actually also missing the case when $x \neq 0$ but $y = 0$, so I’ll have to think about that. (Edit: I think it actually fails there?) Do you have any tips on how I can proceed with Q2? – TuringTester69 Sep 24 '20 at 21:56
  • @TuringTester69 : For Q2: the closed sets are the arbitrary intersections of complements of open sets. So start by looking at the complements of the basic open sets. One that may take some time to see is "start anywhere (including the point 'anywhere'), go left to $-\infty$, pass through ${0}$, continue going left through $\infty$, then stop somewhere (including the point 'somewhere')" is a closed set. One way to see this: the complement of, say $(-1,2)$, meets all of the new neighborhoods of $0$. – Eric Towers Sep 24 '20 at 22:44
  • Brian M. Scott, Could you explain how can we do that? – 00GB Oct 29 '21 at 03:34
  • @00GB : Why would Brian M. Scott see your comment? You haven't @-tagged him to be notified and you haven't posted your comment to his Answer. – Eric Towers Oct 29 '21 at 05:13
  • @EricTowers. Thank you so much to let me know. Sometimes, , I am trying to use @ but and use the first letter but I can not see the name. So, This is why I just wrote the name. – 00GB Oct 29 '21 at 15:14