Arbitrary close and norm-bounded approximations to an element in a Banach space from a dense subset I have an element $e \in E$, where $E$ is a Banach space, and $\lVert e\rVert=l$. 
I also have a subset $Y$ which is dense in $E$.
Does this mean that we can find element $y\in Y$ s.t. $\lVert y - e\rVert \le \frac{l}{2}$ and $\lVert y\rVert \le l$? If so, why?
I have the feeling that a Banach space is like the Euclidean space $\mathbf R^n$ so to speak continuous and we can find elements arbitrary close to an element.
Thanks in advance!
 A: This is a bit fiddly to prove because of the requirement that $\| y \| \leq l$. But I believe it can be done.
Rather than trying to approximate $e$ directly, I'm going to approximate $\frac 3 4 e$. Indeed, this is possible: since $Y$ is dense in $E$, there exists an $y \in Y$ such that
$$ \| y - \tfrac 3 4 e \| < \tfrac l 4.$$
Applying the triangle inequality twice, we have
$$ \| y\| \leq \| y - \tfrac 3 4 e \| + \|\tfrac 3 4 e \| < \tfrac l 4 +\tfrac {3l} 4 = l$$
and
$$ \| y - e \| \leq \| y - \tfrac 3 4 e \| + \| \tfrac 3 4 e - e \| < \tfrac l 4 + \tfrac l 4 = \tfrac l 2.$$

[Added to address point raised in comments]
It is even possible to find a $y \in Y$ such that $\| y - e \| \leq \frac l 2$ and $\| y \| \leq l$ and $y \neq e$.
We would simply choose a $y \in Y$ that approximates $\tfrac 3 4 e$ even closer, for example, we would choose a $y \in Y$ such that
$$ \| y - \tfrac 3 4 e \| < \tfrac l 8.$$
The previous argument still applies, so we still get $\| y - e \| \leq \frac l 2$, and $\| y \| \leq l$.
And $\| y - \tfrac 3 4 e \| < \tfrac l 8$ also tells us that $y$ cannot possibly be equal to $e$.
