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suppose we have the metric space $(S_M,d_\infty)$, where ;

$$S_M:=\{f:\Bbb R \rightarrow\Bbb R \mid f \text{ is continuous and } |f(x)|<M \,\,\forall x \in \Bbb R \}$$

$$d_\infty:=\sup\{|f(x)-g(x)|x \in \Bbb R\}.$$

I want to prove this space is complete.

Here is what I said;

If we have a set of functions $f_n \in S_M$ which are cauchy wrt $d_{\infty}$( henceforth to be referred to as simply d), Then for all $x \in \Bbb R$ $f_n(x)$ is cauchy wrt $d$ and so convergent.

Next we define $f$ to be the pointwise limit of $f_n$. We must show that $f$ is bounded , continuous and that $d(f_n,f)\rightarrow 0,$ as $n \rightarrow \infty$. However we already have that any sequence in this space is bounded and continous so we need only show that $d(f_n,f)\rightarrow 0,$ as $n \rightarrow \infty$.

Take $\varepsilon>0$ and find N s.t. $d(f_n,f_m)<\varepsilon/ 2$ whenever $n,m \geq N$

Since $f_m(x)$ converges to $f(x)$ then there is some $N'$ st $|f_m(x)-f(x)|<\varepsilon/2$ when m$\geq N$

for $m \geq max \{N,N'\}$ we have

$$|f_n(x)-f(x)| \leq |f_n(x)-f_m(x)|+|f_m(x)-f(x)|<\varepsilon/2 + \varepsilon/2 = \varepsilon$$

so as $|f_n(x)-f(x)| < \varepsilon$, we have that $\sup|f_n(x)-f(x)|\leq \varepsilon$

since this holds for all $n \geq N$ we have $lim_{n \rightarrow \infty} \sup_{x \in X} |f_n(x)-f(x)|=0$.

which means that the limit point is contained within the space and so the space is complete.

Am I missing anything important that I should show , any major holes in logic ? I saw an example of a similar question where someone went into proving that $f$ was bounded and continuous but I think that just seems like a lot of extra work given the fact that $f$ must be both of those things by definition if it belongs to $S_M$

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  • $\begingroup$ In the definition of $S_m$, is $M=m$? $\endgroup$ – José Carlos Santos Jun 7 '18 at 22:05
  • $\begingroup$ @JoséCarlosSantos Do you mean is big M in the sets definition equal to the little m that i used when speaking about the sequence of functions ? $\endgroup$ – excalibirr Jun 7 '18 at 22:07
  • $\begingroup$ Yes, that's what I am asking. $\endgroup$ – José Carlos Santos Jun 7 '18 at 22:08
  • $\begingroup$ Then no, I don't think so. The definition of the space is from a past exam paper so maybe that's what they meant although I didn't think that's what they meant and didn't answer it that way. $\endgroup$ – excalibirr Jun 7 '18 at 22:10
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Your space is not complete. Just take the sequence $(f_n)_{n\in\mathbb N}$ of constant functions defined by $f_n(x)=M-\frac1n$. It is a Cauchy sequence, but it is not convergente in $S_M$, since it converges to the constant function $M$, which is outside $S_M$.

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  • $\begingroup$ that makes sense. I had thought my space was the same as this one math.stackexchange.com/questions/347571/… , would you mind explaining how they are different ? or are they the same space but the answer in the link is incorrect ? $\endgroup$ – excalibirr Jun 7 '18 at 22:17
  • $\begingroup$ @exodius Your space is different : it's not the set of all bounded continuous functions (which is complete) but the set of all continuous functions "strictly bounded by a fixed pregiven $M$". $\endgroup$ – Henno Brandsma Jun 7 '18 at 22:20
  • $\begingroup$ @exodius Your space is not the space of all bounded functions. It's a non-closed subspace of that space. If the condition $<M$ gets replaced by $\leqslant M$, then your space will become complete. $\endgroup$ – José Carlos Santos Jun 7 '18 at 22:20
  • $\begingroup$ @JoséCarlosSantos ah okay I see now the distinction. Thank you :) $\endgroup$ – excalibirr Jun 7 '18 at 22:23

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