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67 | 67 |
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68 | 68 | $A = \{(x, 0) : x \in \R\}$. Since this a straight line in $\R^2$, $\partial A = A$ and $A^\circ = \emptyset$. |
69 | 69 |
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70 | | -% TODO: Justify and nice-ify |
71 | | -$B = \{(x, y) : x, y \in \R, x > y, y \ne 0\} = \{(x, y) : x, y \in \R, x > y\} \setminus A$. |
| 70 | +% TODO: Justify |
| 71 | +$B = \{(x, y) : x, y \in \R, x > y, y \ne 0\} = \{(x, y) : x, y \in \R, x > y\} \setminus A$. Therefore \[ |
| 72 | +\partial B = \{(x, x) : x \in \R\} \cup \{(x, 0) : x \in \R, x > 0\} |
| 73 | +\] and $B^\circ = B$. |
72 | 74 |
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73 | | -$\partial B = \{(x, x) : x \in \R\} \cup \{(x, 0) : x \in \R, x > 0\}$. |
74 | | - |
75 | | -$B^\circ = B$. |
76 | | - |
77 | | -$C = \{(x, y) : x, y \in \R, x > y\} \cup A$. |
78 | | - |
79 | | -$\partial C = \{(x, x) : x \in \R\} \cup \{(x, 0) : x \in \R, x < 0\}$. |
80 | | - |
81 | | -$C^\circ = \{(x, y) : x, y \in \R, x > y\}$. |
| 75 | +$C = \{(x, y) : x, y \in \R, x > y\} \cup A$. Therefore \[ |
| 76 | +\partial C = \{(x, x) : x \in \R\} \cup \{(x, 0) : x \in \R, x < 0\} |
| 77 | +\] and \[ |
| 78 | +C^\circ = \{(x, y) : x, y \in \R, x > y\}. |
| 79 | +\] |
82 | 80 |
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83 | 81 | % }}} |
84 | 82 |
|
85 | 83 | % {{{ Q4 |
86 | | -\newquestion{4} |
| 84 | +% \newquestion{4} |
87 | 85 |
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88 | | -\begin{questionbody} |
89 | | -Show that $X$ is Hausdorff if and only if the \textit{diagonal} $\Delta = \{x \times x : x \in X\}$ is closed in $X \times X$. |
90 | | -\end{questionbody} |
| 86 | +% \begin{questionbody} |
| 87 | +% Show that $X$ is Hausdorff if and only if the \textit{diagonal} $\Delta = \{x \times x : x \in X\}$ is closed in $X \times X$. |
| 88 | +% \end{questionbody} |
| 89 | + |
| 90 | +% Answer |
91 | 91 |
|
92 | | -Answer |
93 | 92 | % }}} |
94 | 93 |
|
95 | 94 | % {{{ Q5 |
96 | | -\newquestion{5} |
| 95 | +% \newquestion{5} |
97 | 96 |
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98 | | -\begin{questionbody} |
99 | | -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 $x \in A$. |
100 | | -\end{questionbody} |
| 97 | +% \begin{questionbody} |
| 98 | +% 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 $x \in A$. |
| 99 | +% \end{questionbody} |
101 | 100 |
|
102 | | -% See problem sheet 5, exercise 5 |
| 101 | +% % See problem sheet 5, exercise 5 |
103 | 102 |
|
104 | | -Answer |
| 103 | +% Answer |
105 | 104 |
|
106 | 105 | % }}} |
107 | 106 |
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