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Let $f, g: X \rightarrow \mathbb{R}$ be continuous functions, where ($X, \tau$) is a topological space and $\mathbb{R}$ is given the standard topology.

a)Show that the function $f \cdot g : X \rightarrow \mathbb{R}$,defined by $(f \cdot g)(x) = f(x)g(x)$

is continuous.

b)Let $h: X \setminus \{x \in X | g(x) = 0\}\rightarrow \mathbb{R}$ be defined by $h(x) = \frac{f(x)}{g(x)}$

Show that $h$ is continuous.

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up vote 3 down vote accepted

The central fact is that the operations $$p:\ {\mathbb R}^2\to{\mathbb R},\quad (x,y)\mapsto x\cdot y$$ and similarly $q:\ (x,y)\mapsto {\displaystyle{x\over y}}$ are continuous where defined and that $$h:\ X\to{\mathbb R}^2,\quad x\mapsto\bigl(f(x),g(x)\bigr)$$ is continuous if $f$ and $g$ are continuous.

It follows that $f\cdot g=p\circ h$ is continuous, and similarly for ${\displaystyle{f\over g}}$.

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(a) Given $a \in X$, as $f$ is continuous for all sequencie $(x_k) \subset X$, where $x_k \to a $ implies $f(x_k) \to f(a)$, like wise $g(x_k) \to g(a)$. As $(x_k) \subset \mathbb{R}$ we have $$f(x_k)g(x_k) \to f(a)g(a), \ x_k \to a.$$ Therefore $f.g(x) \to f.g(a), \ x \to a.$

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Well the sequence def of continuity works for metric spaces, here we have topological space. I guess that this is a problem, isnt it? – user10293123 Feb 19 '13 at 11:15
+1 because if one replaces "sequence" with "net" the proof is valid. – user59671 Mar 14 '13 at 11:09

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