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If someone could help explain/hint at what I should do next, I feel like I could do the rest of the problem.

Let $B=\left\{f(t)=C[-1,1]:f(0)=1,|f(t)-1|\leq1,t\in [-1,1]\right\}$

(B is the set of all continuous functions on [-1,1])

We want to prove that $B$ is closed in $C[-1,1]$

$\textbf{What I know}$

We need to take an arbitrary sequence $(y_n)_k$ of functions from $B$ and show that the sequence converges. The we need to show that the limit is actually in the set as well:

(i) show the limit is continuous on [-1,1]

(ii) Satisfies the conditions to be in $B$

I guess I should start by writing lim$|(y_n)_k-y|$ (Where $y$ is the proposed limit), but I am still very confused on what to do next. I ultimately want to answer my own question.

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    Closed in what space? Maybe you want to talk about compactness or completeness. – 2017-02-03
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    $B$ is not closed: the sequence $f_n(x) = |1-x|^n$ is a counterexample. – 2017-02-03
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    Sorry, I have made an edit to closed @edm – 2017-02-03

2 Answers 2

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Let $(X,d)$ be a metric space and let $u\in X.$ Then $\{v\in X:d(u,v)\leq 1\}$ is closed.

Let $1'$ be the function with $1'(x)=1$ for all $x\in [-1,1].$

With the metric $d(f,g)=\sup \{|f(x)-g(x)|: |x|\leq 1\}$ on $C[-1,1],$ let $A_1=\{f\in C[-1,1]: d(f,1')\leq 1\}.$ Then $A_1$ is closed.

Let $A_2=\{f\in C[-1,1]: f(0)=1\}.$ It is easy to see that $A_2$ is closed. The set $B$ in your Q is $A_1\cap A_2,$ which is the intersection of two closed sets. So $B$ is closed.

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    Your idea makes some sense (ideas such as Metric spaces and the use of the 1' function are not things we have covered). My professor told me to try using the fact $lim||y_n-y||\rightarrow 0$ and that $y$ is automatically continuous. Could you maybe give me a hint on where to apply that? – 2017-02-03
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    The $1'$ function is just a constant function. I'm sure you've seen those before..... An arbitrary sequence of members of $B$ need not converge.... The closure of $B$ in this Q is the closure with respect to the metric $d(f,g)=\sup \{|f(x)-g(x)|\}$ on $ C[-1,1]$ so if $f$ is a limit point of $B$ in the space $C[-1,1$] it must be continuous. The fact that $C[-1,1]$ with the $\sup$ metric is a complete metric space is another issue. Note that convergence of $(f_n)_n$ to $f$ in $C[-1,1]$ means uniform convergence. – 2017-02-05
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    I figured it out by using the facts that $(f_n)$'s and limit $f$ converge uniformly and that $f$ is continuous, but will accept your answer to appreciate your time. – 2017-02-14
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$B=\left\{y\in C[-1,1]:f(0)=1,|f(t)-1|\leq 1, t\in C[-1,1]|\right\}$

To show $B$ is closed, we must show that any Cauchy sequence in $B$ has a limit that is also in $B$.

Let $(f_n)$ be a any Cauchy sequence of functions where each $f_n\in B$

By an earlier theorem, there exists a function $f$ such that $\lim(f_n)=f$ and:

(1) $\exists n\in \mathbb{N}: (f_n)$ converges uniformly to $f$ (2) $f$ is continuous on $[-1,1]$

Let $n$ be fixed as in property (1) above:

(i) $|f(t)-1|=|f(t)-f_n(t)+f_n(t)-1|\leq |f(t)-f_n(t)|+|f_n(t)-1|<\epsilon+1$

(ii) Consider $\lim(f_n(0))=\lim(1)=1$, but $\lim(f_n(0))=f(0)$, so $f(0)=1$

By (i), (ii), and (2), $f\in B$