Vtyjkyuk

Problem: Show that the discontinuous and bounded functions set is open in $mathbbB(M,N)$, where $mathbbB(M,N)$ is the set of bounded functions.

My attempt of proof: Let $D_a$ be the set of discontinuous and bounded functions $f:Mrightarrow N$ discontinuous in $a$. So, there exist some $varepsilon >0$ such that for all $delta>0$ we can find an $xin M$ with $d(x,a)<delta$ and $d(f(x),f(a))geq varepsilon$. Now, let $g$ be in $mathbbB(M,N)$ and $d(g,f)<fracvarepsilon3$, then, since
beginalign*
varepsilon
&leq d(f(x),f(a)) \
&leq d(f(x),g(x))+d(g(x),g(a))+d(f(a),g(a)) \
&< fracvarepsilon3+d(g(x),g(a))+fracvarepsilon3
endalign*

Then, $fracvarepsilon3leq d(g(x),g(a))$, that means $gin D_a$. Then, since $f,gin D_a$ and $d(f,g)<fracvarepsilon3$ we can assure that there is a ball $B(f,fracvarepsilon3)subset mathbbB(M,N)$ with $fin D_a$, so $D_a$ is open in $mathbbB(M,N)$.

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 $B(f,fracvarepsilon3)subset mathbbB(M,N)$ with $fin D_a$” 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 $D_a subset mathbbB(M,N)$ of bounded functions which are discontinuous at this specific point $a$ is open.
  You still need to conclude that the set $D subseteq mathbbB(M,N)$ of bounded functions which are discontinuous at some point is open.
  Hint:

  
  
  

  
  

  We have that $D = bigcup_a in M D_a$.