3
$\begingroup$

Consider a space of smooth functions $C^{\infty}[a,b]$ and a set $$\tau=\left\{B(f,\varepsilon_0,\varepsilon_1...\varepsilon_r):f\in C^{\infty}[a,b],r\in\mathbb{N}\right\} $$ where $f$ is arbitrary function and $r$ is an arbitrary number from $\mathbb{N}$ and $$B(f,\varepsilon_0,\varepsilon_1...\varepsilon_r)=\left\{h\in C^{\infty}[a,b]: |f^{(k)}(x)-h^{(k)}(x)|<\varepsilon_{k} \forall x\in[a,b] \forall k\in \overline{0,r}\right\} $$

The aim is to prove that $\tau$ is a base of topology on $C^{\infty}[a,b]$. We can use a criterium (it consists of 2 statements): $$1. \forall h\in C^{\infty}[a,b] \quad \exists B(f,\varepsilon_0,\varepsilon_1...\varepsilon_r):h\in B(f,\varepsilon_0,\varepsilon_1...\varepsilon_r) $$ This is obvious if we take $B(h,\varepsilon_0,\varepsilon_1...\varepsilon_r)$.

$$2.\forall B(f,\varepsilon_0,\varepsilon_1...\varepsilon_r),B(g,\mu_0,\mu_1...\mu_r):B(f,\varepsilon_0,\varepsilon_1...\varepsilon_r)\bigcap B(g,\mu_0,\mu_1...\mu_r)=C\ne \varnothing $$ $$\forall h\in C \quad \exists W\in \tau:h\in W\subset C$$ Here I have stuck. I can show 2nd statement only for each $k\in \overline{0,r}$. For example, I can find $W$ for $k=0$, but there is no guarantee that $W$ can satisfy the rest of requirements. I will be grateful if you help me with this stuff.

$\endgroup$
3
  • $\begingroup$ To be more concrete, just take $B(h,1)$ for the first one. $\endgroup$ Oct 15, 2014 at 20:24
  • $\begingroup$ For the second, they need not have the same $r$, in general they will have $r_1$ and $r_2$. $\endgroup$ Oct 15, 2014 at 20:24
  • $\begingroup$ But otherwise how can we define an intersection with different $r$? $\endgroup$
    – cool
    Oct 15, 2014 at 20:57

1 Answer 1

1
+50
$\begingroup$

Let $h\in B(f,\epsilon_0,...,\epsilon_r)\cap B(g,\mu_0,...,\mu_s)$ and suppose $r\leq s$. Let $h$ be a function in the intersection. We shall write $||q||=\sup_{x\in [a,b]} q(x)$ for the sup norm. For $i\leq r$ let $\eta_i$ be such that $||h^{(i)}-f^{(i)}||+\eta_i<\epsilon_i,||h^{(i)}-g^{(i)}||<\mu_i$. For $i>r,$ just let $\eta_i$ be such that $||h^{(i)}-g^{(i)}||<\mu_i$. I claim that $B=B(h,\eta_0,...,\eta_s)$ is contained in the intersection. Let $k\in B$. Then $k^{(i)}$ is within $\eta_i$ of $h^{(i)}$, so by the triangle inequality within $\mu_i$ of $g^{(i)}$ and similarly within $\epsilon_i$ of $f^{(i)}$ whenever $i\leq r$.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks, @Kevin Carlson. This is the proof I was seaching for. $\endgroup$
    – cool
    Oct 18, 2014 at 17:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .