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If $f : X \to \mathbb{R}$ is continuous,

I want to show that $(cf)(x) = cf(x)$ is continuous, where $c$ is a constant.

Attempt: If $f$ is continuous, then we want to show that the inverse image of every open set in $\mathbb{R}$ is an open set of $X$. Choose an open interval in $\mathbb{R}$.

Thats as far as I got. :(

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I think you mean you want to show $(cf)(x)=cf(x)$ is continuous. – Grumpy Parsnip Feb 26 '13 at 23:46
Sorry about that. Jim, you are right. Thats what I am trying to solve. – user64013 Feb 26 '13 at 23:48
more generally, show that the composition of two continuous functions is continuous. or, in particular here, $c^{-1}U$ is an open interval in $\mathbb{R}$ when $U$ is an open interval in $\mathbb{R}$ – yoyo Feb 26 '13 at 23:55
Let $g:\mathbb{R}\to\mathbb{R}$ be $g(y) = cy$. Your function is $g\circ f$. – Willie Wong Feb 27 '13 at 0:19
If you’ve not yet done so, you should certainly prove that the composition of continuous functions is continuous. It’s an easy argument, and the fact itself is enormously useful. – Brian M. Scott Feb 27 '13 at 6:22

Let $g:\mathbb R\to \mathbb R$ be $g(y)=cy$. Your function is $g\circ f$. - Willie Wong

The continuity of $g$ is shown by directly verifying that the preimage of any open interval is an open interval. (The case $c=0$ is somewhat exceptional and can be dealt with separately: constant maps are easily seen to be continuous.)

You will likely have other opportunities to turn arithmetical operations into composition: e.g., the product $fg$ is the composition of the map $(f,g):X\to\mathbb R^2$ with the multiplication map $h:\mathbb R^2\to \mathbb R$ defined by $h(u,v)=uv$. To show that $fg$ is continuous, it suffices to check the continuity of $h$.

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An alternative answer, that does not require you to prove the composition of continuous functions is continuous:

Let $U\subset\mathbb{R}$ be open. In fact we may assume $U=(a,b)$ by taking the open balls as a base for $\mathbb{R}$.


$$ (cf)^{-1}(U) = \{ x\in X: \exists y\in(a,b): (cf)(x) = y\} = \left\{x\in X:\exists y\in(a,b): f(x)=\frac{y}{c}\right\} $$ $$ =\left\{x\in X: \exists z\in \left(\frac{a}{c},\frac{b}{c}\right): f(x)=z\right\} $$

and this last set is precisely $$f^{-1}\left\{\left(\frac{a}{c},\frac{b}{c}\right)\right\},$$ which is open since $f$ is continuous.

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Of course, technically I should justify dividing by $c$, but clearly $c=0$ is a trivial case. – Gyu Eun Lee Jul 14 '13 at 0:59

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