Is $C[0,1]$ open in $B[0,1]$? Suppose that


*

*$B[0,1]$ := set of all bounded functions on $[0,1]$ equipped with the topology induced by the sup-norm

*$C[0,1]$ :=set of all continuous functions on $[0,1]$.
Is $C[0,1]$ open in $B[0,1]$?
 A: Hint: Let $f$ be a continuous function. Find a discontinuous function in any ball centered at $f$ (you only need to change the value of $f$ at one point). Why does this show $C[0,1]$ not open?
A: Indeed you can show more.


*

*As you probably know, a uniform limit of continuous functions is continuous.  Once you have convinced yourself that convergence in the sup norm is the same as uniform convergence, you'll see this implies that $C[0,1]$ is closed in $B[0,1]$.

*Now why does this imply that $C[0,1]$ is not open?

*Verify that $B[0,1]$ (or indeed any topological vector space) is connected (indeed, path connected, and the paths can be very simple...)

*In a connected topological space $X$, the only sets which are both open and closed are $\emptyset$ and $X$.  Put another way, a proper nonempty subset which is closed can never be open.
A: If $C[0,1]$ is open in $B[0,1]$ then it is an open vector subspace of $B[0,1]$.
Lemma: $(V,||.||)$ be an NLS. $Y$ is open subspace of $V$ then $Y=V$. 
Hence $C[0,1]$ cannot be open in $B[0,1]$
