5
$\begingroup$

Here is Prob. 12, Chap. 4 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

A uniformly continuous function of a uniformly continuous function is uniformly continuous.

State this more precisely and prove it.

Here is my effort:

Theorem:** Let $\left(X, d_X\right)$, $\left(Y, d_Y \right)$, and $\left( Z, d_Z \right)$ be metric spaces, let $f$ be a uniformly continuous mapping of $X$ into $Y$, let $g$ be a uniformly continuous mapping of $f(X)$ into $Z$, and let $h = g \circ f$. Then $h$ is a uniformly continuous mapping of $X$ into $Z$.

Proof:** Let $\varepsilon$ be a given real number such that $\varepsilon > 0$. Since $g$ is a unifromly continuous mapping of $f(X)$ into $Z$, we can find a real number $\eta > 0$ such that $$\tag{1} d_Z \left( g \left( y_1 \right), g \left( y_2 \right) \right) < \varepsilon$$ for any points $y_1$ and $y_2$ in $f(X)$ for which $$\tag{2} d_Y \left( y_1, y_2 \right) < \eta.$$ Now as $f$ is a uniformly continuous mapping of $X$ into $Y$, so, corresponding to the real number $\eta > 0$ in particular, we can find a real number $\delta > 0$ such that $$ \tag{3} d_Y \left( f \left(x_1 \right), f \left( x_2 \right) \right) < \eta$$ for any points $x_1$ and $x_2$ in $X$ for which $$ \tag{4} d_X \left( x_1, x_2 \right) < \delta.$$ So, we can conclude from (1), (2), (3), (4) above that, for any points $x_1$ and $x_2$ in $X$ which satisfy $$d_X \left( x_1, x_2 \right) < \delta,$$ the following is true. $$ d_Z \left( h \left(x_1 \right), h\left( x_2 \right) \right) = d_Z \left( g\left( f\left( x_1 \right) \right), g\left( f\left( x_2 \right) \right) \right) < \varepsilon, $$ from which it follows that $h = g \circ f$ is a uniformly continuous mapping of $X$ into $Z$.

Have I managed to get the statement of the theorem right? If so, then is my proof correct?

  • 0
    Looks good to me. – 2017-01-30
  • 0
    Looks good to me too. +1. – 2017-01-30
  • 0
    Nice! +1 from me as well. Maybe post your proof as an answer! – 2017-01-30
  • 0
    @UmbertoP. can you please also have a look at the following question of mine? http://math.stackexchange.com/questions/2120085/prob-11-chap-4-in-baby-rudin-uniform-continuous-extension-from-a-dense-subse – 2017-01-30
  • 0
    @positrón0802 thank you. I would be grateful if you could also take time going through my question at http://math.stackexchange.com/questions/2120085/prob-11-chap-4-in-baby-rudin-uniform-continuous-extension-from-a-dense-subse. – 2017-01-30
  • 0
    @TheCount so nice of you. Can you please also have a look at this post? Here's the link. http://math.stackexchange.com/questions/2120085/prob-11-chap-4-in-baby-rudin-uniform-continuous-extension-from-a-dense-subse – 2017-01-30
  • 0
    Don't post a comment like that @SaaqibMahmuud three times in a row to different people; they're all still visible to us. – 2017-01-30
  • 0
    @SeanRoberson sorry! I wanted to make sure each one does get to read my request, in case they've moved elsewhere. – 2017-01-30
  • 0
    They won't get moved; everybody seems them, and anybody can see any other questions you may have from the main page or your profile. – 2017-01-30
  • 0
    It isn't a big deal, Saaqib, don't worry about it. Just don't make a habit of it. :) – 2017-01-30

1 Answers 1

1

This looks perfect. You have the right conceptual idea, as well: simply shrink the ball in $X$ until its image "fits" inside the ball in $Y$, then send it to $Z$. Nicely done.

  • 0
    can you please also have a look at the question of mine at the following link? http://math.stackexchange.com/questions/2120085/prob-11-chap-4-in-baby-rudin-uniform-continuous-extension-from-a-dense-subse – 2017-01-30