Problem: Show that the discontinuous and bounded functions set is open in $\mathbb{B}(M,N)$, where $\mathbb{B}(M,N)$ is the set of bounded functions.

My attempt of proof: Let $D_{a}$ be the set of discontinuous and bounded functions $f:M\rightarrow N$ discontinuous in $a$. So, there exist some $\varepsilon >0$ such that for all $\delta>0$ we can find an $x\in M$ with $d(x,a)<\delta$ and $d(f(x),f(a))\geq \varepsilon$. Now, let $g$ be in $\mathbb{B}(M,N)$ and $d(g,f)<\frac{\varepsilon}{3}$, then, since \begin{align*} \varepsilon &\leq d(f(x),f(a)) \\ &\leq d(f(x),g(x))+d(g(x),g(a))+d(f(a),g(a)) \\ &< \frac{\varepsilon}{3}+d(g(x),g(a))+\frac{\varepsilon}{3} \end{align*}

Then, $\frac{\varepsilon}{3}\leq d(g(x),g(a))$, that means $g\in D_{a}$. Then, since $f,g\in D_{a}$ and $d(f,g)<\frac{\varepsilon}{3}$ we can assure that there is a ball $B(f,\frac{\varepsilon}{3})\subset \mathbb{B}(M,N)$ with $f\in D_{a}$, so $D_{a}$ is open in $\mathbb{B}(M,N)$.

Question to you guys: Is this proof correct?, how could you improve it? Thanks for your help! :)


Your proof is overall corrent, but I think it can be improved in terms of clarity.

  • You should make it clear that you fix $f \in D_a$ and show that $D_a$ contains an open ball around $f$.

    • One the one hand this helps the reader to understand what you’re about to do.
    • On the other hand it makes clear that the appearing constants (in this case $\varepsilon$) depend on the choice $f$, which can help preventing mistakes in the proof.
  • Instead of “there is a ball $B(f,\frac{\varepsilon}{3})\subset \mathbb{B}(M,N)$ with $f\in D_{a}$” what you need—and what you have shown—is that “the ball $B(f, \varepsilon/3)$ is completely contained in $D_a$”.

  • So far you have only shown that for every $a \in M$ the subset $D_a \subset \mathbb{B}(M,N)$ of bounded functions which are discontinuous at this specific point $a$ is open. You still need to conclude that the set $D \subseteq \mathbb{B}(M,N)$ of bounded functions which are discontinuous at some point is open. Hint:

    We have that $D = \bigcup_{a \in M} D_a$.

  • 1
    $\begingroup$ Thanks so much for your answer, then, to be strict, since $D=\bigcup D_{a}$, and $D_{a}$ is open in $\mathbb{B}(M,N)$ and the arbitrary union of open set is open, then D is open in $\mathbb{B}(M,N)$, right? $\endgroup$ – user1trill Sep 8 '18 at 21:29
  • 1
    $\begingroup$ Yes, this works. $\endgroup$ – Jendrik Stelzner Sep 8 '18 at 22:06

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