Topology: An Inquiry-Based Approach

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

\(a_n = (2,n)\) in \((\R^2, d_M)\)

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 \(n \in \Z^+\text{,}\) let \(B_n = B\left(x;m+\frac{1}{n}\right)\text{.}\) Why must \(B_n \cap A \neq \emptyset\) for each \(n \in \Z^+\text{?}\)

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 \((Y,d')\) be a subspace of \((X,d)\text{.}\) Let \(a_1\text{,}\) \(a_2\text{,}\) \(\ldots\) be a sequence of points in \(Y\) and let \(a \in Y\text{.}\) Prove that if \(\lim_n a_n = a\) in \((Y,d')\text{,}\) then \(\lim_n a_n = a\) in \((X,d)\text{.}\)

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

Show that \(\lim ka_n = k \lim a_n\) 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 \(|a_n| \leq M\) for all \(n \in \Z^+\text{.}\)

Show that \(\lim a_nb_n = \lim a_n \ \lim b_n\text{.}\)

If \(b_n \neq 0\) for every \(n\) and \(\lim b_n \neq 0\text{,}\) 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 \(g(x) \neq 0\) for every \(x \in \R\text{.}\) Define the function \(\frac{f}{g}\) from \(\R\) to \(\R\) by \(\frac{f}{g}(x) = \frac{f(x)}{g(x)}\) for every \(x \in \R\text{.}\) 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 \(\R\) are given the Euclidean topology.

Modify the function \(f\) to construct a new function \(g: \R \to \R\) 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 \(h: \R \to \R\) that is continuous at any given finite number of points?

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

If \((a_n)\) is a sequence in \((\R, d_E)\) with \(a_{n+1} \lt a_n\) for each \(n \in \Z^+\) and the set \(\{a_n\}\) is bounded below, then \(\inf \{a_n \mid n \in \Z^+\}\) 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 \(r \gt 0\text{,}\) then there is a sequence in \(A\) that converges to \(a\text{.}\)

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

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 \(N \in \Z^+\) such that \(a_n = a\) for all \(n \geq N\text{.}\))