Topology: An Inquiry-Based Approach

$a_n=1+\frac{1}{n}$ in $(\mathbb{R},d_E)$

$a_n=(2,n)$ in $({\mathbb{R}}^2,d_M)$

Show that if $A$ is bounded above, then there is a sequence $(a_n)$ in $A$ such that $lim{a}_n=sup(A)\text{.}$

Show that if $A$ is bounded below, then there is a sequence $(a_n)$ in $A$ such that $lim{a}_n=inf(A)\text{.}$

Are the limits from (a) or (b) necessarily in $A\text{?}$ Explain.

For each $n\in {\mathbb{Z}}^{+}\text{,}$ let $B_n=B(x;m+\frac{1}{n})\text{.}$ Why must $B_n\cap A\ne \mathrm{\varnothing }$ for each $n\in {\mathbb{Z}}^{+}\text{?}$

Let $a_n\in B_n\cap A$ for each $n\text{.}$ What property does this sequence have? Explain how we have just proved the following theorem.

Let $(Y,d^{\prime })$ be a subspace of $(X,d)\text{.}$ Let $a_1\text{,}$ $a_2\text{,}$ $\dots$ be a sequence of points in $Y$ and let $a\in Y\text{.}$ Prove that if $\underset{n}{lim}{a}_n=a$ in $(Y,d^{\prime })\text{,}$ then $\underset{n}{lim}{a}_n=a$ in $(X,d)\text{.}$

Show that the converse of part (a) is false by considering the subspace $(\mathbb{Q},d_{\mathbb{Q}})$ (the rational numbers) of $(R,d)\text{.}$ Let $a_1\text{,}$ $a_2\text{,}$ $\dots$ be a sequence of rational numbers such that $\underset{n}{lim}{a}_n=\sqrt{2}\text{.}$ Prove that , given $ϵ>0\text{,}$ there is a positive integer $N$ such that for $n,m>N\text{,}$ $|a_n-a_m|<ϵ\text{.}$ Does the sequence $a_1\text{,}$ $a_2\text{,}$ $\dots$ converge when considered to be a sequence of points in $\mathbb{Q}\text{?}$

Show that $limk{a}_n=klim{a}_n$ for any real number $k\text{.}$

Show that $lim(a_n+b_n)=lim{a}_n+lim{b}_n\text{.}$

Show that the sequence $(a_n)$ is bounded. That is, show that there is a positive real number $M$ such that $|a_n|\le M$ for all $n\in {\mathbb{Z}}^{+}\text{.}$

Show that $lim{a}_n{b}_n=lim{a}_{n}lim{b}_n\text{.}$

If $b_n\ne 0$ for every $n$ and $lim{b}_n\ne 0\text{,}$ show that $lim\frac{a_n}{b_n}=\frac{lim{a}_n}{lim{b}_n}\text{.}$

Prove that $fg$ is a continuous function.

Assume that $g(x)\ne 0$ for every $x\in \mathbb{R}\text{.}$ Define the function $\frac{f}{g}$ from $\mathbb{R}$ to $\mathbb{R}$ by $\frac{f}{g}(x)=\frac{f(x)}{g(x)}$ for every $x\in \mathbb{R}\text{.}$ Use the definition of continuity to prove that $\frac{f}{g}$ is a continuous function.

Show that $f$ is continuous at exactly one point. Assume that both copies of $\mathbb{R}$ are given the Euclidean topology.

Modify the function $f$ to construct a new function $g:\mathbb{R}\to \mathbb{R}$ such that $g$ is continuous at exactly the numbers $0$ and $1\text{.}$ Prove your result. Can you see how to extend this to construct a function $h:\mathbb{R}\to \mathbb{R}$ that is continuous at any given finite number of points?

Let $0\le a<1\text{.}$ Show that the sequence $(a_n)$ where $a_n=a^n$ converges to $0$ in $(\mathbb{R},d_E)\text{.}$

If $(a_n)$ is a sequence in $(\mathbb{R},d_E)$ with $a_{n+1}<a_n$ for each $n\in {\mathbb{Z}}^{+}$ and the set $\{a_n\}$ is bounded below, then $inf\{a_n\mid n\in {\mathbb{Z}}^{+}\}$ is the limit of the sequence $(a_n)\text{.}$

Let $X$ be a metric space and $A$ a nonempty subset of $X\text{.}$ If $a\in X$ and if $B(a,r)$ in $X$ contains a point of $A$ for every $r>0\text{,}$ then there is a sequence in $A$ that converges to $a\text{.}$

Let $R$ be a nonempty subset of $\mathbb{R}$ that is bounded above and below. If $S$ is a nonempty subset of $\mathbb{R}$ and $x\le y$ for all $x\in S$ and for all $y\in R\text{,}$ then $sup(S)\le inf(R)\text{.}$

The only convergent sequences in a metric space $(X,d)$ with discrete metric $d$ are the sequences that are eventually constant. (A sequence $(a_n)$ in a metric space $X$ is eventually constant if there is an element $a\in X$ and an $N\in {\mathbb{Z}}^{+}$ such that $a_n=a$ for all $n\ge N\text{.}$)