Exercise TG.1

Proof. Suppose that $X$ is a set with the discrete topology and $C$ is a finite point set. Then $X-C$ is clearly still a subset of $X$ and so is open since $X$ is discrete. This shows by definition that $C$ is closed. In fact by this same argument any subset of $X$ is both open and closed. □

Proof. It suffices to show that the subset of $X\times Y$ containing a single arbitrary element is open, since clearly any other subset is the union of such single-element open subsets and is therefore also open by the definition of a topology. So consider any $(x,y)\in X\times Y$ and the subset $\left\{(x,y)\right\}\subset X\times Y$. Then clearly $\left\{(x,y)\right\}=\left\{x\right\}\times \left\{y\right\}$, which is a basis element of $X\times Y$ and therefore open by the definition of a product topology since both $\left\{x\right\}$ and $\left\{y\right\}$ are open in $X$ and $Y$, respectively, since they are discrete. □

Proof. This is fairly obvious since, for any open subset $V$ of $Y$, of course $f^{-1}(V)$ is a subset of $X$ and so is open since $X$ is discrete. □

Proof. First we must show that $(\mathbb{Z},+)$ is even a group. Clearly $a+b$ is an integer when $a$ and $b$ are so that the closure axiom is satisfied. Also, we know that integer addition is associative. We clearly have that $0\in \mathbb{Z}$ and that $a+0=a$ for any $a\in \mathbb{Z}$ so that $0$ is the identity element of $(\mathbb{Z},+)$. Lastly, for any $a\in \mathbb{Z}$, we have that $-a\in \mathbb{Z}$ and that $a+(-a)=a-a=0$ so that clearly $-a$ is the inverse of $a$. This shows that $(\mathbb{Z},+)$ is in fact a group.

To show that it is a topological group, we first note that $\mathbb{Z}$ clearly has the discrete topology when considered both an order topology or as a subspace of $ℝ$, for similar reason as discussed in Example 3 of §14. Thus $\mathbb{Z}$ satisfies the $T_1$ axiom by Lemma 2 since it is discrete. Also $X\times X$ is the discrete topology by Lemma 3 . Thus the function $f$ defined by $f(x\times y)=x+y^{-1}=x+(-y)=x-y$ is a function from $X\times X$ to $X$, so that it follows that $f$ is continuous by Lemma 4 since $X\times X$ is discrete. Hence $(\mathbb{Z},+)$ is a topological group by Exercise .1. □

Proof. Similarly to part (a), clearly $(ℝ,+)$ is a group with identity element $0$ and inverse element $-x$ for any $x\in ℝ$. However, this time the topology is no longer discrete. Of course we know that $ℝ$ satisfies the $T_1$ axiom. Now consider the function $f(x_{}\times y_{})=x+y^{-1}=x+(-y)=x-y$ for any $x,y\in ℝ$. Consider also any basis element $B=(a,b)\in ℝ$, where here we are of course using the order topology basis. Then we clearly have

Clearly this is the region in ${ℝ}^2$ between the lines $y=x-b$ and $y=x-a$, which is obviously an open set in ${ℝ}^2$. This shows that $f$ is continuous since $B$ was an arbitrary basis element, so that $(ℝ,+)$ is a topological group by Exercise .1. □

Proof. First, clearly $ℝp$ satisfies the $T_1$ axiom since $ℝ$ does. Next we note that for any $x,y\in ℝp$ we have that $x\cdot y$ is also positive so that $x\cdot y\in ℝp$ as well, which shows the closure property of a group. Also, clearly $1\in ℝp$ is the identity element of multiplication, and the inverse element is $x^{-1}=1∕x$ for any $x\in ℝp$, noting that this is defined since $x>0$, and that $1∕x>0$ so that $x^{-1}=1∕x\in ℝp$. Lastly, we know that multiplication is associative on the reals (and therefore also on $ℝp$), which completes the check that $(ℝp,\cdot )$ is in fact a group.

As before, define the function $f:ℝp\times ℝp\to ℝp$ by $f(x_{}\times y_{})=x\cdot y^{-1}=x\cdot 1∕y=x∕y$. Consider the order topology basis of $ℝ$ and consider any basis element $B$ of the subspace $ℝp$ so that $B=ℝp\cap (a,b)$ for some $a,b\in ℝ$ where $a<b$ by Lemma 16.1. Now, if $a\le 0$ then clearly $B=(0,b)$, and we have that

noting that $0<b$ so that $x∕b$ is defined. Obviously this is the region in $ℝp\times ℝp$ ($ℝp\times ℝp$ being the upper right quadrant of ${ℝ}^2$ that does not include either axis) above the line $y=x∕b$, which is easy to show is open in $ℝp\times ℝp$.

On the other hand, if $a>0$ than $B=(a,b)$ so that

which is clearly the region of $ℝp\times ℝp$ between the lines $y=x∕b$ and $y=x∕a$. It is easy to see that again this is an open subset of $ℝp\times ℝp$, which shows that $f$ is continuous either way. This in turn proves that $(ℝp,\cdot )$ is a topological space, again by Exercise .1. □

Proof. Topologies on the complex plane $ℂ$ have not really been discussed, but $ℂ$ is usually defined as $ℝ\times ℝ$ having the usual product topology. Then of course $S^1$ is the unit circle in $ℂ$. We know that $ℝ$ is Hausdorff so that $ℂ=ℝ\times ℝ$ is as well by Theorem 17.11. Then, again by Theorem 17.11, $S^1$ is Hausdorff since it is a subspace of $ℂ$, and so it also satisfies the $T_1$ axiom.

While perhaps not immediately obvious, it is easy to show that $S^1$ is closed under multiplication. If $z,w\in S^1$ then $\left|z\right|=\left|w\right|=1$ so that $\left|z\cdot w\right|=\left|z\right|\cdot \left|w\right|=1\cdot 1=1$ by familiar rules of complex analysis so that $z\cdot w\in S^1$ as well. Clearly $1\in S^1$ is the identity element where the inverse element of $z\in S^1$ is $1∕z$, noting that $\left|1∕z\right|=1∕\left|z\right|=1∕1=1$ since $z\in S^1$, and so $1∕z\in S^1$. We also note that $\left|0\right|=0$, and hence $0\notin S^1$ so that the inverse $1∕z$ is always defined. Lastly, we know that multiplication is associative within $ℂ$ and therefore also within $S^1$. This shows that $(S^1,\cdot )$ satisfies all of the group axioms.

To rigorously show that $S^1$ is a topological group is actually quite tedious so we shall omit some details. Suppose that $U$ is open in $S^1$ and that $z\times w\in f^{-1}(U)$ so that $f(z\times w)\in U$. Now, clearly the unit circle in $ℂ=ℝ\times ℝ$ is the set $S^1=\left\{e^i\mid \in ℝ\right\}$ so that we can express $z=e^i$ and $w=e^{i\varphi }$ for some $,\varphi \in ℝ$. We then have that

While tedious to show rigorously, it follows from the fact that $U$ is open in $S^1$ that there is an $𝜖>0$ where $f(z\times w)\in A_{-\varphi ,𝜖}\subset U$, where we define

noting that of course $A_{\alpha ,𝜖}\subset S^1$. Now consider $A_{,𝜖∕2}$ and $A_{\varphi ,𝜖∕2}$, which are both clearly open in $S^1$ and noting that clearly $z\in A_{,𝜖∕2}$ and $w\in A_{\varphi ,𝜖∕2}$. For any $z^{\prime }=e^{i^{\prime }}\in A_{,𝜖∕2}$ and $w^{\prime }=e^{i{\varphi }^{\prime }}\in A_{\varphi ,𝜖∕2}$ we then have that

Hence

so that $f(z^{\prime }\times w^{\prime })=e^{i{(}^{\prime }-{\varphi }^{\prime })}\in A_{-\varphi ,𝜖}\subset U$. Thus $z^{\prime }\times w^{\prime }\in f^{-1}(U)$ so that $z\times w\in A_{,𝜖∕2}\times A_{\varphi ,𝜖∕2}\subset f^{-1}(U)$ since $z^{\prime }$ and $w^{\prime }$ were arbitrary. We also have that $A_{,𝜖∕2}\times A_{\varphi ,𝜖∕2}$ is open in $S^1\times S^1$ since both $A_{,𝜖∕2}$ and $A_{\varphi ,𝜖∕2}$ are open in $S^1$. Since $z\times w$ was an arbitrary element of $f^{-1}(U)$, this shows that $f^{-1}(U)$ is open in $S^1\times S^1$, which in turn shows that $f$ is continuous by definition. Thus by Exercise .1 we have that $S^1$ is a topological group. □

Proof. First, from linear algebra we know that the matrix product of two nonsingular $n$ by $n$ matrices is another nonsingular $n$ by $n$ matrix, so that $GL(n)$ is closed under matrix multiplication. Clearly the identity matrix is the identity element of $GL(n)$, while the inverse matrix $A^{-1}$ is the inverse element of the matrix $A\in GL(n)$, noting that this inverse matrix exists since $A$ is nonsingular. Lastly, we know that matrix multiplication is associative, which suffices to show that $(GL(n),\cdot )$ is a group.

To show that it is a topological group takes more work. To begin, we note that of course ${ℝ}^{n^2}$ is Hausdorff and so satisfies the $T_1$ axiom. Thus so does $GL(n)$ since ${ℝ}^{n^2}$ gives it its topology. Next, we denote a vector in ${ℝ}^n$ by $x_n=x_1\times \cdots \times x_n$, using the subscript on the vector itself to indicate its dimension. We show that the function $s_n:{ℝ}^n\to ℝ$ defined by

is continuous for all $n\in {\mathbb{Z}}_{+}$, which we show by induction. First, for $n=1$, we clearly have that $s_n$ is simply the identity function from $ℝ$ to $ℝ$, which is clearly continuous. Now suppose that $s_n$ is continuous. Define $g:{ℝ}^{n+1}\to {ℝ}^n$ by

which is continuous by Theorem 19.6 since we know that each ${\pi }_i$ is continuous.

Then also $s_n\circ g$ is continuous by Theorem 18.2 part (c). It then follows that the function $h:{ℝ}^{n+1}\to {ℝ}^2$ defined by $h(x_{n+1})=(s_n\circ g)(x_{n+1})\times {\pi }_{n+1}(x_{n+1})$ is continuous by Theorem 18.4 since both $s_n\circ g$ and ${\pi }_{n+1}$ are continuous. Lastly we then have that $k:{ℝ}^{n+1}\to ℝ$ defined by $+\circ h$ is continuous by Theorem 18.2 part (c), where of course $+$ is the usual addition operation from ${ℝ}^2$ to $ℝ$, which we showed is continuous in Exercise 21.12. Now we claim that $k=s_{n+1}$. For any $x_{n+1}\in {ℝ}^{n+1}$ we have

This completes the induction since we have shown that $k=s_{n+1}$ is continuous.

Next we show that the function $p_{ij}:{ℝ}^n\times {ℝ}^n\to ℝ$ defined by $p_{ij}(x_n\times y_n)=x_i{y}_j$ is continuous, where of course $ij\in \left\{1,\dots ,n\right\}$. Define the function $g_{ij}:{ℝ}^n\times {ℝ}^n\to {ℝ}^2$ by $g_{ij}={\pi }_i\times {\pi }_j$ as in Exercise 18.10, which we know is continuous by that exercise since the coordinate functions are continuous. Then the function $\cdot \circ g_{ij}$ from ${ℝ}^n\times {ℝ}^n$ to $ℝ$ is also continuous by Theorem 18.2 part (c), where of course $\cdot$ is the normal multiplication operation from ${ℝ}^2$ to $ℝ$, which we know is continuous from Exercise 12.12. However, for any $x_n,y_n\in {ℝ}^n$, we have

so that $\cdot \circ g_{ij}=p_{ij}$ is continuous, which shows the desired result.

Now, by definition, each matrix component of the resultant matrix in matrix multiplication on $GL(n)$ is a sum of products, where each product involves a term from each of the matrices, and the sum has $n$ terms going across a row of the first matrix and a column of the second. Thus each component is a composition of the sum function $s_n:{ℝ}^n\to ℝ$ with a mapping $f$ from ${ℝ}^{n^2}\times {ℝ}^{n^2}\to {ℝ}^n$, where each element in ${ℝ}^n$ of this mapping is a product function $p_{ij}$. Since have shown above that each $p_{ij}$ is continuous, it follows from Theorem 19.6 that the mapping $f$ is also continuous. Hence the composition $s_n\circ f$, i.e. the matrix component function, is also continuous by Theorem 18.2 part (c) since we have also shown above that $s_n$ is continuous. Since each component function is continuous, it again follows from Theorem 19.6 that the overall matrix multiplication mapping from ${ℝ}^{n^2}\times {ℝ}^{n^2}\to {ℝ}^{n^2}$ is continuous.

Regarding the inverse element function, we recall from linear algebra that in the inverse of a matrix $A\in GL(n)$ is

where $adj(A)$ is the adjugate matrix of $A$ and $\left|A\right|$ is the determinant of $A$, noting that this is nonzero since $A$ is nonsingular. Now, the determinant is a sum of products so that the function $g:GL(N)\to ℝ$ defined by $g(A)=\left|A\right|$ is continuous by the same arguments as above for matrix multiplication. Likewise each element of the adjugate matrix is a sum of products as well so that the function $f_{ij}:GL(A)\to ℝ$ defined by $f_{ij}(A)=adj{(A)}_{ij}$, i.e. the $i$th row and $j$th column component of the adjugate matrix, is also continuous.

Then clearly the corresponding component of the inverse matrix is the function $h_{ij}:GL(A)\to ℝ$ defined by $h_{ij}(A)=f_{ij}(A)∕g(A)$. Since both $f_{ij}$ and $g$ are continuous (and again noting that $g$ is always nonzero), then their quotient $h_{ij}$ is also continuous by Exercise 21.12. Hence, since each component $h_{ij}$ of the inverse matrix is continuous, it follows that the inversion operation as a whole is continuous by Theorem 19.6 as above, considering the matrices as elements of ${ℝ}^{n^2}$. Since both multiplication and inversion are continuous, this shows that $GL(n)$ is a topological group by definition. □