# Basic (Metric) Topology

Let $(X,d)$ be a metric space.

Definition. A subset $U\subset X$ is said to be open if $\forall x\in U$ $\exists\epsilon>0$ s.t. $B(x,\epsilon)\subset U$.

If $U\subset X$ is open then $U$ can be expressed as union of open balls $B(x,\epsilon)$. Hence, the set of all open balls in $X$, $\mathcal{B}=\{B(x,\epsilon): x\in X,\ \epsilon>0\}$ form a basis for a topology (a metric topology, the topology induced by the metric $d$) on $X$. Those who have not studied topology before may simply understand it as the set of all open sets in $X$.

Definition. A subset $F\subset X$ is said to be closed if its complement, $F^c=X\setminus F$ is open in $X$.

The following is the definition of a continuous function that you are familiar with from calculus. The definition is written in terms of metrics.

Definition. Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. A mapping $T:X\longrightarrow Y$ is said to be continuous at $x_0\in X$ if $\forall\epsilon>0$ $\exists\delta>0$ s.t $d_Y(Tx,Tx_0)<\epsilon$ whenever $d_X(x,x_0)<\delta$.

$T$ is said to be continuous if it is continuous at every point of $X$.

The above definition can be generalized in terms of open sets as follows.

Theorem. A mapping $T: (X,d_X)\longrightarrow(Y,d_Y)$ is continuous if and only if $\forall$ open set $U$ in $Y$, $T^{-1}U$ is open in $X$.

Proof. (Only if, $\Rightarrow$) Suppose that $T:X\longrightarrow Y$ is continuous. Let $U$ be open in $Y$. Then we show that $T^{-1}U$ is open in $X$. Let $x_0\in T^{-1}U$. Then $Tx_0\in U$. Since $U$ is open in $Y$, $\exists\epsilon>0$ s.t. $B(Tx_0,\epsilon)\subset U$. By the continuity of $T$, for this $\epsilon>0$ $\exists\delta>0$ s.t. whenever $d(x,x_0)<\delta$, $d(Tx,Tx_0)<\epsilon$. This means that
$$TB(x_0,\delta)\subset B(Tx_0,\epsilon)\subset U\Longrightarrow B(x_0,\delta)\subset T^{-1}(TB(x_0,\delta))\subset T^{-1}U.$$ Hence, $T^{-1}U$ is open in $X$.

(If, $\Leftarrow$) Suppose that $\forall$ open set $U$ in $Y$, $T^{-1}U$ is open in $X$. We show that $T$ is continuous. Let $x_0\in X$ and let $\epsilon>0$ be given. Then $B(Tx_0,\epsilon)$ is open in $Y$. So by the assumption, $x_0\in T^{-1}B(Tx_0,\epsilon)$ is open in $X$. This means that $\exists\delta>0$ s.t.
$$B(x_0,\delta)\subset T^{-1}B(Tx_0,\epsilon)\Longrightarrow TB(x_0,\delta)\subset T(T^{-1}B(Tx_0,\epsilon))\subset B(Tx_0,\epsilon).$$ This is equivalent to saying that $\exists\delta>0$ s.t. whenever $d(x,x_0)<\delta$, $d(Tx,Tx_0)<\epsilon$. That is, $T$ is continuous at $x_0$. Since the choice $x_0\in X$ was arbitrary, the proof is complete.

Let $A\subset X$. $x\in X$ is called an accumulation point or a limit point of $A$ if $\forall$ open set $U(x)$ in $X$, $(U(x)-\{x\})\cap A\ne\emptyset$. Here the notation $U(x)$ means that it contains $x$. The set of all accumulation points of $A$ is denoted by $A’$ and is called the derived set of $A$. $\bar A:=A\cup A’$ is called the closure of $A$. $\bar A$ is the smallest closed set containing $A$.

Theorem. Let $A\subset X$. Then $x\in\bar A$ if and only if $\forall$ open set $U(x)$, $U(x)\cap A\ne\emptyset$.

Definition. $D\subset X$ is said to be dense if $\bar D=X$. This means that $\forall$ open set $U$ in $X$, $U\cap D\ne\emptyset$.

Definition. $X$ is said to be separable if it has a countable dense subset.

Examples. The real line $\mathbb{R}$ is separable. The complex plane $\mathbb{C}$ is also separable.

Theorem. The space $\ell^\infty$ is not separable.

Proof. Let $y=(\eta_1,\eta_2,\eta_3,\cdots)$ be a sequence of zeros and ones. Then $y\in\ell^\infty$. We can then associate $y$ with the binary representation
$$\hat y=\frac{\eta_1}{2}+\frac{\eta_2}{2^2}+\frac{\eta_3}{2^3}+\cdots\in [0,1].$$ Each $\hat y\in [0,1]$ has a binary representation and different $\hat y$’s have different binary representations. So, there are uncountably many sequences of zeros and ones. If $y$ and $z$ are sequences of zeros and ones and $y\ne z$, then $d(y,z)=1$. This means that for any two distinct sequences $y$ and $z$ of zeros and ones, $B\left(y,\frac{1}{3}\right)\cap B\left(z,\frac{1}{3}\right)=\emptyset$. Let $A$ be a dense subset of $\ell^\infty$. Then for each sequence $y$ of zeros and ones, $B\left(y,\frac{1}{3}\right)$ has at least one element of $A$. This means that $A$ cannot be countable.

Theorem. The space $\ell^p$ with $1\leq p<\infty$ is separable.

Proof. Let $A$ be the set of all sequences $y$ of the form
$$y=(\eta_1,\eta_2,\cdots,\eta_n,0,0,\cdots,0),$$ where $n$ is a positive integer and the $\eta_j$’s are rational. For each $n=1,2,\cdots$, the number of sequences of the form $y=(\eta_1,\eta_2,\cdots,\eta_n,0,0,\cdots,0)$ is the same as the number of functions from $\{1,2,3,\cdots,n\}$ to $\mathbb{Q}$, the set of all rational numbers. $\mathbb{Q}$ has the cardinality $\aleph_0$ and so the number is $\aleph_0^n=\aleph_0$. The cardinality of $A$ is then $\aleph_0\cdot\aleph_0=\aleph_0$ i.e. $A$ is countable. Now we show that $A$ is dense in $\ell^p$. Let $x=(\xi_j)\in\ell^p$. Let $\epsilon>0$ be given. Since $\displaystyle\sum_{j=1}^\infty|\xi_j|^p<\infty$, $\exists$ a positive integer $N$ s.t. $\displaystyle\sum_{j=N+1}^\infty|\xi_j|^p<\frac{\epsilon^p}{2}$. Since rationals are dense in $\mathbb{R}$, one can find $y=(\eta_1,\eta_2,\cdots,\eta_N,0,0,\cdots)\in A$ s.t. $\displaystyle\sum_{j=1}^N|\xi_j-\eta_j|^p<\frac{\epsilon^p}{2}$. Hence,
$$[d(x,y)]^p=\sum_{j=1}^N|\xi_j-\eta_j|^p+\sum_{j=N+1}^\infty|\xi_j|^p<\epsilon^p,$$
i.e. $d(x,y)<\epsilon$. This means that $y\in B(x,\epsilon)\cap A\ne\emptyset$. This completes the proof.