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Give up on Topologies Ass 3
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second-year/MA260-Norms-Metrics-and-Topologies/Ass-3/main.tex

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$A = \{(x, 0) : x \in \R\}$. Since this a straight line in $\R^2$, $\partial A = A$ and $A^\circ = \emptyset$.
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% TODO: Justify and nice-ify
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$B = \{(x, y) : x, y \in \R, x > y, y \ne 0\} = \{(x, y) : x, y \in \R, x > y\} \setminus A$.
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% TODO: Justify
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$B = \{(x, y) : x, y \in \R, x > y, y \ne 0\} = \{(x, y) : x, y \in \R, x > y\} \setminus A$. Therefore \[
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\partial B = \{(x, x) : x \in \R\} \cup \{(x, 0) : x \in \R, x > 0\}
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\] and $B^\circ = B$.
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$\partial B = \{(x, x) : x \in \R\} \cup \{(x, 0) : x \in \R, x > 0\}$.
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$B^\circ = B$.
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$C = \{(x, y) : x, y \in \R, x > y\} \cup A$.
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$\partial C = \{(x, x) : x \in \R\} \cup \{(x, 0) : x \in \R, x < 0\}$.
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$C^\circ = \{(x, y) : x, y \in \R, x > y\}$.
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$C = \{(x, y) : x, y \in \R, x > y\} \cup A$. Therefore \[
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\partial C = \{(x, x) : x \in \R\} \cup \{(x, 0) : x \in \R, x < 0\}
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\] and \[
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C^\circ = \{(x, y) : x, y \in \R, x > y\}.
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\]
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% }}}
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% {{{ Q4
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\newquestion{4}
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% \newquestion{4}
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\begin{questionbody}
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Show that $X$ is Hausdorff if and only if the \textit{diagonal} $\Delta = \{x \times x : x \in X\}$ is closed in $X \times X$.
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\end{questionbody}
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% \begin{questionbody}
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% Show that $X$ is Hausdorff if and only if the \textit{diagonal} $\Delta = \{x \times x : x \in X\}$ is closed in $X \times X$.
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% \end{questionbody}
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% Answer
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Answer
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% }}}
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% {{{ Q5
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\newquestion{5}
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% \newquestion{5}
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\begin{questionbody}
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Let $A \subset X$ and let $f : A \to Y$ be continuous. Suppose $Y$ is Hausdorff. Show that there is at most one continuous function $g : \ol A \to Y$ such that $g(x) = f(x)$ for all $x \in A$.
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\end{questionbody}
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% \begin{questionbody}
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% Let $A \subset X$ and let $f : A \to Y$ be continuous. Suppose $Y$ is Hausdorff. Show that there is at most one continuous function $g : \ol A \to Y$ such that $g(x) = f(x)$ for all $x \in A$.
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% \end{questionbody}
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% See problem sheet 5, exercise 5
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% % See problem sheet 5, exercise 5
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Answer
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% Answer
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% }}}
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