Weakest topology that makes the interval $(a,b]$ open

general-topologygroup-theorytopological-groupsweak-topology

I have these two questions:

1) Let $G$ be $\mathbb{R}$ with addition as the group operation and with the weakest topology that makes each interval of the form $(a,b]$ open. Show that $x\rightarrow -x$ is not continuous and hence $G$ is not a topological group.

I know that weakest mean that that topology is the smallest topology, in other words, it is contained by all other topologies that make each interval of the form $(a,b]$ open. What I don't understand is what it means for the topology to make the interval $(a,b]$ open. Is that this topology the collection of arbritrary union of sets of the form $\{(a,b):a,b \in \mathbb{R}\}$? This is especially confusing because every element of the topology is called open.

I would like to know then what is the weakest topology that makes each interval of the form $(a,b]$ open.

2) Let $G$ be $\mathbb{R}$ with addition as the group operation and with the topology for which the open sets are either empty or have countable complement (check that this is a topology on $G$). Show that $(x,y)\rightarrow xy$ is continuous if either one of the entries is held fixed but that is not jointly continuous. Hence $G$ is not a topological group.

This question is somewhat similar in nature to the first one. I have the same issue. I don't really know what that topology looks like.

I took a look at this. That makese sense but I don't know how to extend that to find the topologies for these problems.

I know these questions might be very basic but I am new to topology, so I haven't quite wrapped my head around it yet. Any help would be appreciated.

Best Answer

The collection of all subsets of $\Bbb R$ of the form $(a,b]$ is a base for a topology on $\Bbb R$: you can check that it satisfies the necessary conditions to be a base. That topology consists of the empty set and all sets that are unions of arbitrary families of intervals of the form $(a,b]$. For instance, $\{x\in\Bbb R:x\le 0\}$ is open in this topology, because it is the union of the intervals $(-n,0]$ for $n\in\Bbb Z^+$. $(0,1)$ is also open in this topology: $(0,1)=\bigcup_{x\in(0,1)}(0,x]$. This topology is sometimes called the upper-limit topology on $\Bbb R$. Any topology that includes all of the sets of the form $(a,b]$ must contain all unions of such sets, so it must contain the upper-limit topology. This means that the upper-limit topology is the weakest (coarsest) topology making all of those intervals open: any other topology that does so contains all of the open sets in the upper-limit topology and others besides.

One way to show that the map $f:\Bbb R\to\Bbb R:x\mapsto -x$ is not continuous in this topology is to find an open set $U$ in this topology whose inverse image $f^{-1}[U]$ under $f$ is not open in this topology. Try it with $U=(-2,-1]$, which by definition is open in this topology: find $f^{-1}[U]$ and explain why it isn’t open in this topology. HINT: Find a point $p\in f^{-1}[U]$ that has no open nbhd contained in $U$.

The topology in the second question is the co-countable topology on $\Bbb R$ and is completely described in the problem statement: it consists of the empty sets and the subsets $U$ of $\Bbb R$ such that $\Bbb R\setminus U$ is countable. (Note that countable includes finite, so that $\Bbb R\setminus\{0\}$, for instance, is open in the co-countable topology: its complement is the finite (and therefore countable) set $\{0\}$.) We don’t have to build the open sets from some family of basic open sets as we did in the first problem: we already know exactly what the open sets are.

Now you’re asked to consider the map $f:\Bbb R\times\Bbb R\to\Bbb R:\langle x,y\rangle\mapsto xy$, so the first thing to do is to figure out what the product topology on $\Bbb R\times\Bbb R$ is when each factor has the co-countable topology. Remember that the product topology has a base consisting of the products of open sets in the two factors; these are the sets of the form $U\times V$, where $U$ and $V$ are co-countable subsets of $\Bbb R$. That is, there are countable subsets $C$ and $D$ of $\Bbb R$ such that $U=\Bbb R\setminus C$ and $V=\Bbb R\setminus D$. As a non-trivial example, $\Bbb Q$, the set of rationals, is countable, so $\Bbb R\setminus\Bbb Q$, the set of irrationals, is open in the co-countable topology. So is $\Bbb R\setminus\Bbb Z$, the set of non-integers. The Cartesian product of these sets, $(\Bbb R\setminus\Bbb Q)\times(\Bbb R\setminus\Bbb Z)$, is therefore an open set in $\Bbb R\times\Bbb R$ in the product topology; it consists of all points $\langle x,y\rangle$ in the plane such that $x$ is irrational and $y$ is not an integer.

To finish the problem you must show three things.

  1. You must show that $f$ is not continuous. HINT: Let $U=\Bbb R\setminus\{1\}$; $U$ is open in the co-countable topology. Figure out what $f^{-1}[U]$ is as a subset of the plane and show that it is not open in the product topology on $\Bbb R\times\Bbb R$.
  2. You must show that if we fix $p\in\Bbb R$ and let $f_p:\Bbb R\to\Bbb R:\langle p,y\rangle\mapsto py$, the map $f_p$ is continuous with respect to the co-countable topology. To do this, let $U$ be an arbitrary open set in $\Bbb R$, and show that $f_p^{-1}[U]$ is open in the domain.
  3. You must show that if we fix $p\in\Bbb R$ and let $f^p:\Bbb R\to\Bbb R:\langle x,p\rangle\mapsto xp$, the map $f^p$ is also continuous. This is essentially the same as the second task.

(We say that $f$ is separately continuous, meaning that it is continuous in each variable separately when the other is held fixed, but not jointly continuous, i.e., not continuous as a function of two variables.)

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