I wish to prove that continuous image of a closed bounded subset of $\mathbb{R}$ is closed and bounded (the function is $\mathbb{R}\rightarrow\mathbb{R} $). However, I do not have the equivalence to compactness in my system. My definition of a closed set is that its complement is open. I also know that a closed set contains all the limits of Cauchy sequences in it.
Is there a way I can do this, or is the compactness argument necessary?
Thanks for your help!
Best Answer
You'll first need the lemma:
Then if $f:\mathbb R\to\mathbb R$ is continuous, and $X\subset[a,b]$ is closed, then if $f$ is not bounded on $X$, find $x_n\in X$ so that $|f(x_n)|>n$. Then $\{x_n\}$ has a Cauchy sub-sequence, which must converge to a value in $X$ since $X$ is closed. But then $f(x_n)\to x$. Is that possible?
To prove that $f(X)$ is closed, you do likewise. If $y$ is a limit point of $f(X)$, let $x_1,x_2,\dots,x_n,\dots\in X$ so that $f(x_n)\to y$. Then take a Cauchy sub-sequence of $\{x_1,\dots,x_n,\dots\}$.
The proof of the lemma:
Given an infinite sequence $\{x_1,x_2,\dots\}\subseteq [a_0,b_0]$. Set $n_0=1$.
Given $n_k,a_k,b_k$ with infinitely many elements of the $x_i$ in $[a_k,b_k]$, we split the interval in half.
If there are infinitely many $x_i\leq \frac{a_k+b_k}{2}$, we let $a_{k+1}=a_k, b_{k+1}=\frac{a_k+b_k}{2}$.
Otherwise, choose $a_{k+1}=\frac{a_k+b_k}{2}$ and $b_{k+1}=b_k$.
In either case, there are infinitely many $x_i$ in $[a_{k+1},b_{k+1}]$. Choose $n_{k+1}$ so that $n_{k+1}> n_k$ and $x_{n_{k+1}}\in [a_{k+1},b_{k+1}]$.
By induction, since $[a_{k+1},b_{k+1}]\subset [a_k,b_k]$ we have that if $i\geq k$ then $x_{n_i}\in [a_k,b_k]$.
Now $b_k-a_k=\frac{b_0-a_0}{2^k}$ so if $i,j\geq k$, then $|x_{n_i}-x_{n_j}|\leq \frac{b_0-a_0}{2^k}$. This shows that $\{x_{n_i}\}$ is Cauchy.