So I'm trying to prove that the set of polynomials in C([a,b],R) is not open. I understand the definition of an open set, but I'm wondering how to find a point that is not contained in the interior if it's the set of polynomials.

Also, Can a subset of a metric space ever be both open and dense? I'm thinking it can, only if the interior is empty. ?

Would love help with these! Thanks in advance!

  • $\begingroup$ A subset of a metric space can be both open and dense. Consider $\mathbb{R}\setminus\{0\}$ in $\mathbb{R}$. $\endgroup$ – Michael Albanese Sep 24 '14 at 0:17
  • $\begingroup$ What metric on $C([a,b],\mathbb{R})$ are you using? $\endgroup$ – JimmyK4542 Sep 24 '14 at 0:18

The "usual" metric on $C([a,b], \mathbb{R})$ is $$d(f,g) = \sup_{x \in [a,b]} |f(x) - g(x)|$$ Please let me know if that is not the metric you are assuming.

Let $P$ denote the set of polynomials in $C([a,b], \mathbb{R})$.

If $P$ is open, then for any function $f \in P$, there must be some $\epsilon$-neighborhood of $f$ which contains only elements of $P$. In other words, there must be some $\epsilon > 0$ such that, if $g \in C([a,b], \mathbb{R})$ and $d(f,g) < \epsilon$, then $g$ is a polynomial.

We will show that $P$ is not open by showing that for any $f \in P$ and any $\epsilon > 0$, we can find a non-polynomial in $C([a,b], \mathbb{R})$ for which $d(f,g) < \epsilon$. This will show that $f$ is not an interior point of $P$.

Let $f \in P$ and let $\epsilon > 0$. Then define $g(x) = f(x) + (\epsilon/2) \sin(x)$. Note that $g \in C([a,b], \mathbb{R})$ since $g$ is the sum of two continuous functions. Also note, that $g$ is not a polynomial: if it were, then $(\epsilon/2) \sin(x) = g(x) - f(x)$ would be the difference of two polynomials, hence a polynomial. But this is not the case, because $\sin(x)$ is not the zero function and has infinitely many zeros, whereas a nonzero polynomial can only have finitely many zeros.

Now we compute the distance between $f$ and $g$: $$\begin{align} d(f,g) &= \sup_x |g(x) - f(x)|\\ &= \sup_x |(\epsilon/2) \sin(x)|\\ &= (\epsilon/2) \sup_x |\sin(x)|\\ &= \epsilon/2 \end{align}$$

Therefore, $d(f,g) < \epsilon$ and we have found the desired contradiction.

Note that this proof shows that $P$ contains no interior points, which is a stronger statement than "$P$ is not open", which only requires us to show that $P$ contains at least one non-interior point.

  • $\begingroup$ This is pretty helpful. Is there a chance you could go through this in more detail? Why did you pick this metric and how did you know to pick it? and the last sentence is hard for me to digest... I'm having quite trouble understanding this. $\endgroup$ – user146296 Sep 24 '14 at 1:09
  • $\begingroup$ @user146296: I added some more detail above. The metric I used is the "usual" one on $C([a,b], \mathbb{R})$. However, only you can tell us if that is the metric assumed for this problem. $\endgroup$ – Bungo Sep 24 '14 at 1:22
  • $\begingroup$ sometimes we define other metrics, so I was unsure which I should assume. I guess the question is not clear to me. When it's asking if set of polynomials f(x) is not an open set I took that as (there exists a point p that has a radius epsilon, s.t. the ball has points that are not in the set [a,b]) is it equivalent to say there is no ball with radius epsilon around f(x) containing all polynomials? $\endgroup$ – user146296 Sep 24 '14 at 1:27
  • $\begingroup$ Thank you so much!! You have helped my understanding so much! I really appreciate you taking the time to walk me through this! $\endgroup$ – user146296 Sep 24 '14 at 1:31

Hint: Find a sequence of non-polynomials $f_n$ such that $d(f_n,0) \to 0$.

This will show that the $0$-polynomial is not an interior point of the subset of polynomials.


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