A 'weird' topology

Question

I've got some questions about the following situation, and some proof-verification requests.

Let $\mathcal{T}$ be the smalles topology on $\mathbb{R}$ with the property that $$f:(\mathbb{R},\mathcal{T})\rightarrow(\mathbb{R},\mathcal{T}_{Eucl}), f(x)=x^2$$ is continuous.

Describe a basis of $(\mathbb{R},\mathcal{T})$ and show that any $U\in\mathcal{T}$ has the property that $$-x\in U, \forall x\in U.$$ My attempt: A basis of a topology is a family $\mathcal{B}$ of opens of X with the property that any open $U\subset X$ can be written as a union of opens that belong to $\mathcal{B}$. So $$\mathcal{B}=\{B(x;\epsilon):B(-x;\epsilon)\in \mathcal{B}, x\in\mathbb{R},\epsilon >0\}.$$ In this case, $-x\in U$, $\forall x\in U$. Is this basis correct?

Find the closure and the interior of $(-1,2)$ in $(\mathbb{R},\mathcal{T})$.

My attempt: $x\in \overline{(-1,2)} \Leftrightarrow \forall U\in \mathcal{B}_x, U\cap (-1,2) \neq \emptyset$, with $\mathcal{B}_x$ the basis of neigborhoods of $x$. So the closure of $(-1,2)$ is not only $[-1,2]$, but it is the union $[-1,2]\cup [-2,1] = [-2,2]$. We also know $x\in \text{Int}((-1,2)) \Leftrightarrow \exists U\in \mathcal{B}_x$ s.t. $ U\subset (-1,2)$. This shows the interior of $(-1,2)$ has to be $(-2,2)$. Note that, for example $\{-2\}\not\in (-1,2)$, but if $\{-2\}\in U$, then $\{2\}\in U$, so it has to be in the interior. Am I thinking right?

Next, I thought $(\mathbb{R},\mathcal{T})$ is not Hausdorff. I wrote a prove for this, I am almost sure this is correct.

Is $[-1,1]$, with the topology induced from $\mathcal{T}$, compact? Connected?

My attempt: I think $[-1,1]$ is compact, but I'm not sure how to prove this. I know, $[-1,1]\subset (-2,2)$ and $(-2,2)\cap [-1,1]=[-1,1]$, but this is just an example. This is not enough to prove right? Can someone help me with this? I think the interval is connected. My proof: Assume $[-1,1]$ is not connected, so $\exists U,V$ such that $[-1,1]=U\cup V$. Since $U=[-1,1]-V$, U must be closed in $[-1,1]$. Hence, as a limit of points in $U$, $R:=\sup U$ must belong to $U$. We claim that $R=1$. If not, we find an interval $(-R-\epsilon, R+\epsilon)\subset U$ and then $R+\frac{1}{2}\epsilon$ and $-R-\frac{1}{2}\epsilon$ would be two elements in $U$, strictly greater than $\sup U$. this is impossible, so $-1,1\in U$. But exactly the same argument shows that $-1,1\in V$, and this contradicts the fact that $U\cap V = \emptyset$.

Is [-1,1) compact?

I tried to figure this out, but I failed. Someone who can explain this to me?

Does there exist a metric space $X$ and a finite group $\Gamma$ acting on $X$ such that $X/\Gamma$ is homeomorphic to $(\mathbb{R},\mathcal{T})$?

I also failed to answer this question. I even have no clue how to start this exercise.

I know I am asking a lot, but it would really help me if I know I did this right or how to do it better! I really hope someone would take time to answer this!! :)

Answer 1

Some extended HINTS: The members of $\mathscr{T}$ are precisely the sets $f^{-1}[U]$ such that $U\in\mathscr{T}_{Eucl}$.

• Check that if $\mathscr{B}_{Eucl}$ is a base for $\mathscr{T}_{Eucl}$, then $\mathscr{B}=\left\{f^{-1}[B]:B\in\mathscr{B}_{Eucl}\right\}$ is a base for $\mathscr{R}$.

Now take for $\mathscr{B}_{Eucl}$ the set of open intervals in $\Bbb R$. Let $(a,b)\in\mathscr{B}_{Eucl}$.

• Show that if $a\ge 0$, then $f^{-1}[(a,b)]=(-b,-a)\cup(a,b)$.
• Show that if $a<0<b$, then $f^{-1}[(a,b)=(-b,b)$.
• Show that if $b\le 0$, then $f^{-1}[(a,b)]=\varnothing$, so we can ignore these intervals.

For $A\subseteq\Bbb R$, let $-A=\{-a:a\in A\}$. If $A=-A$, say that $A$ is symmetric.

• Show that $\mathscr{T}=\{U\in\mathscr{T}_{Eucl}:U=-U\}$, the family of symmetric Euclidean-open sets.

When you get to here, you’ve essentially answered the first boldface question.

For the second boldface question, recall that the interior of a set $A$ is the largest open set contained in $A$; $(-2,2)$ is not a subset of $(-1,2)$, so it cannot be the interior of $(-1,2)$. What is the largest symmetric Euclidean-open set contained in $(-1,2)$? As for the closure, it is indeed $[-2,2]$, though your explanation isn’t really very clear.

You are correct in thinking that the new topology is not Hausdorff: if $n\ne 0$, it is never possible to find disjoint open sets containing $x$ and $-x$, respectively. This follows immediately from the fact that the members of $\mathscr{T}$ are all symmetric.

You are also correct in thinking that $[-1,1]$ is compact in the new topology. This is an immediate consequence of the fact that $\mathscr{T}\subseteq\mathscr{T}_{Eucl}$. Suppose that $\mathscr{U}$ is an open cover of $[-1,1]$ in the new topology. Then $\mathscr{U}$ is also an open cover of $[-1,1]$ in the Euclidean topology, so ... ?

You can use the same basic idea to show that $[-1,1]$ is connected in the new topology. If not, there are $U,V\in\mathscr{T}$ such that $U\cap[-1,1]\ne\varnothing\ne V\cap[-1,1]$, $U\cap V\cap[-1,1]=\varnothing$, and $[-1,1]\subseteq U\cup V$. In other words, $U\cap[-1,1]$ and $V\cap[-1,1]$ are relatively open, complementary subsets of $[-1,1]$, so they form a separation of $[-1,1]$. But $\mathscr{T}\subseteq\mathscr{T}_{Eucl}$, so $U$ and $V$ are also open in the usual topology, and hence ... ?

For the compactness of $[-1,1)$:

• Show that $[-1,0]$ is compact in the new topology.
• Then show that $\mathscr{U}\subseteq\mathscr{T}$ covers $[-1,1)$ if and only if it covers $[-1,0]$.

For the last question, let $\Gamma=\{0,1\}$ with multiplication as the group operation, and let $X=\Bbb R$. For $g\in\Gamma$ and $x\in\Bbb R$ I’ll write $g\cdot x$ for the action of $g$ on $x$. By the definition of group action you know that $0\cdot x=x$ for each $x\in\Bbb R$. Find a way to define $1\cdot x$ for $x\in\Bbb R$ so that $X/\Gamma$ is homeomorphic to $\langle\Bbb R,\mathscr{T}\rangle$.