Discuss the convergence of \(f_n(t)=t^n\) in (C[0, 1], \(d_1\)) and (C[0, 1], \(d_{\infty}\)).
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Every metric space (X, d) is Hausdorff: For distinct x, y ∈ X, there exists r > 0 such that \(B_d(x,r)\cap B_d(y,r)=\emptyset\). In particular, limit of a convergent sequence is unique.
(Co-finite Topology) We declare that a subset \(U\) of \(\mathbb{R}\) is open iff either \(U=\emptyset \) or \(\mathbb{R} \backslash U\)is finite. Show that R with this “topology” is not Hausdorff.
A subset \(U\) of a metric space X is closed if the complement X \\(U\) is open. By a neighbourhood of a point, we mean an open set containing that point. A point x ∈ X is a limit point of \(U\)if every non-empty neighbourhood of x contains a point of U. (This definition differs from that given in Munkres). The set \( \overline{U}\) is the collection of all limit points of \(U\).
What are the limit points of bidisc in \(\mathbb{C^2}\) ?
Let (X, d) be a metric space and let \(U\) be a subset of X. Show that \(x\in \overline{U}\), iff for every \(x\in \overline{U}\), there exists a convergent sequence \(\{x_n \} \subseteq U\) such that \(Lim_{n \rightarrow\infty}x_n=x\).
Describe all topologies on a 2-point set. Give five topologies on a 3-point set
Let \((X,\Omega)\) be a topological space and let \(U\) be a subset of \(X\). Suppose for every \(x\in U\) there exists \(U_x\in \Omega\) such that \(x \in U_x \subseteq \Omega\). Show that \(U\) belongs to \(\Omega\).
(Co-countable Topology) For a set \(X\), define \(\Omega\) to be the collection of subsets \(U\) of \(X\) such that either \(U=\emptyset\) or \(X \backslash U\) is countable. Show that \(\Omega\) is a topology on \(X\).
Let \(\Omega\) be the collection of subsets \(U\) of \(X:=\mathbb{R}\) such that either \(X \backslash U= \emptyset\) or \(X\backslash U\) is infinite. Show that \(\Omega\) is not a topology on \(X\).
Show that the usual topology is finer than the co-finite topology on \(\mathbb{R}\).
Show that the usual topology and co-countable topology on \(\mathbb{R}\) are not comparable.
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