Some Proofs
Table of contents
Lecture 2, Example 20
Let $(X,d)$ be a metric space, and let $x\in X$. Then, for any $\epsilon>0$, $B(x,\epsilon)$ is open in $X$. In fact, this ball is sometimes referred to as an open ball .
Proof: Let $y\in B(x,\epsilon)$. If $x = y$ then this is automatically true, just take $\epsilon’ = \frac{\epsilon}{2}$. Suppose that $y\neq 0$, and let $r = \epsilon - d(x,y)>0$. We want to show that $B(y,r) \subset B(x,\epsilon).$ Let $z\in B(y,r).$ Then \(d(x,z) \leq d(x,y) + d(y,z) < d(x,y) + r = \epsilon.\)
Therefore, $B(y,r) \subset B(x,\epsilon)$, and thus $B(x,\epsilon)$ is an open set.
Lecture 2, Proposition 25
Let ${x_n}$ be a sequence in $\mathbb{R}$. Then, ${x_n}$ is convergent (and converges to $x$) if and only if $\forall \epsilon>0$, all but finitely many terms in ${x_n}$ are in $(x-\epsilon, x+\epsilon)$.
Proof: Given $x_n\to x$, given $\epsilon>0$ there exists an $N$ such that for all $n\geq N$, $\vert x_n-x\vert < \epsilon$. Therefore, for all $n\geq N$, $x_n \in (x-\epsilon, x+\epsilon)$. For the other direction, fix arbitrary $\epsilon >0$ and consider $(x-\epsilon, x+\epsilon)$. Given that all but finitely many terms in ${x_n}$ are in $(x-\epsilon, x+\epsilon)$, there exists an $M$ such that for all $n\geq M$, $x_n \in (x-\epsilon, x+\epsilon) = B_\epsilon(x)$. Therefore, $x_n$ is convergent.
Lecture 2, Theorem 29
Let $(X,d_X)$ and $(Y, d_Y)$ be metric spaces. Then, $f:X\to Y$ is continuous at $c\in X$ if and only if for every sequence ${x_n}$ in $X$ with $x_n \to c$, we have that $f(x_n) \to f(c)$.
Proof: Suppose that $f$ is continuous at $c$. Let ${x_n}$ be a sequence in $X$ converging to $c$. Given $\epsilon>0$, there exists a $\delta>0$ such that $d_X(x,c) <\delta \implies d_Y(f(x),f(c))<\epsilon$. Given $x_n\to c$, there exists an $N$ such that for all $n\geq N$, $d_X(x_n, c) <\delta$. Therefore, $d_Y(f(x_n), f(c))<\epsilon$. Thus, $f(x_n) \to f(c)$.
Suppose that $f$ is not continuous at $c$. Let $\epsilon>0.$ Then, for all $n\in \mathbb{N}$, there exists an $x_n$ such that $d(x_n,c) < \frac{1}{n}$ but $d_Y(f(x_n),f(c))\geq \epsilon$. Then, $x_n\to c$ but $f(x_n)$ does not converge to $f(c).$