Cauchy condition for functions

Solution 1:

Suppose that $(x_n)$ converges to $a$. By the Bolzano-Weierstrass theorem, we know that the sequence $(f(x_n))$ must have a monotone subsequence $(f(x_{n_j}))$.

Further, $(f(x_{n_j}))$ must be bounded: taking $\epsilon=1$, there exists $\delta>0$ so that $\lvert x-a\rvert<\delta$ and $\lvert y-a\rvert<\delta$ implies $\lvert f(x)-f(y)\rvert<1$; in particular, for all $x\in(a-\delta,a+\delta)$ we have $\lvert f(x)-f(a+\frac{\delta}{2})\rvert<1$, which implies $$ \lvert f(x)\rvert<\lvert f(a+\tfrac{\delta}{2})\rvert+1\text{ for all }x\in(a-\delta,a+\delta); $$ since $x_{n_j}$ is eventually contained in $(a-\delta,a+\delta)$, the sequence $(f(x_{n_j}))$ is then bounded.

So, $(f(x_{n_j}))$ is bounded and monotone, and therefore converges to some $L$. We claim that $(f(x_n))$ must converge to $L$ as well.

Let $\epsilon>0$ be given. By assumption, there exists $\delta>0$ such that $0<\lvert x-a\rvert<\delta$ and $0<\lvert y-a\rvert<\delta$ implies $\lvert f(x)-f(y)\rvert<\frac{\epsilon}{2}$. Because $x_n\rightarrow a$, there exists $N\in\mathbb{N}$ so that $n>N$ implies $\lvert x_n-a\rvert<\delta$. Because $x_{n_j}\rightarrow a$ and $f(x_{n_j})\rightarrow L$, there exists $J\in\mathbb{N}$ such that $\lvert x_{n_J}-a\rvert<\delta$ and $\lvert f(x_{n_J})-L\rvert<\frac{\epsilon}{2}$.

Then for $n>N$, $$ \lvert f(x_n)-L\rvert\leq\lvert f(x_n)-f(x_{n_J})\rvert+\lvert f(x_{n_J})-L\rvert<\epsilon. $$ So, $f(x_n)\rightarrow L$, as claimed.

Solution 2:

Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces and $f:X \setminus a \to Y$ ($a$ could be in the domain of $f$, but doesn't need to be).
Let $a \in X$ be a limit point of X. Then there is at least one sequence $(a_i)$ of points of X distinct from $a$ converging to $a$, and the sequence is in the domain of $f$.
Let $Y$ be complete so that all Cauchy sequences have a limit point in Y.
Note that for a real function $f: \mathbb R \to \mathbb R$ with the modulus metric $|.|$ these conditions are satisfied, as all points are limit points and $\mathbb R $ is complete.

$f$ having a limit $L$ at $a$ is defined as the function $g$ being continuous at $a$ where $g(a) = L$ and $g(x \ne a) = f(x)$.

In a metric space continuity is equivalent to sequential continuity, i.e. "$g$ is continuous at $a$" is equivalent to "for all sequences $(a_i)$ (of which we have at least one) which converge to $a$ then $g(a_i)$ converges to $g(a)$".

Proof:

(1) We take that given $\epsilon$ there exists $\delta$ such that for $d_X(a, x_1), d_X(a, x_2) < \delta$ then $d_Y(f(x_1), f(x_2)) < \epsilon$.

Let $A = (a_i)$ be a sequence in $X$ converging to $a$.
Then there is $N_A$ such that for $i > N_A$ we have $d_X(a, a_i) < \delta$.
Then let $i, j > N_A$, so by (1), $d_Y(f(a_i), f(a_j)) < \epsilon$ which is the condition that $(f(a_i))$ is a Cauchy sequence in $Y$.
Since $Y$ is complete then the sequence $(f(a_i))$ has a limit in $Y$ say $L$ and we define $g(a) = L$.

It remains to show that for any other sequence $(b_i)$ which converges to $a$ then $(f(b_i))$ converges to the same limit $L$.
Let $N_B$ be such that for $k > N_B$ we have $d_X(a, b_k) < \delta$
Since $(f(a_i))$ converges to $ L$ we can find $N$ such that for $i > N$ we have $d_Y(L, f(a_i)) < \epsilon$
Then let $M = $max$(N, N_A, N_B)$ and for $i, k > N$ we have $d_Y(f(a_i), f(b_k)) < \epsilon$ and $d_Y(L, f(a_i)) < \epsilon$
So by triangle inequality, $d_Y(L, f(b_i)) \le d_Y(L, f(a_i)) + d_Y(f(a_i), f(b_k)) = 2\epsilon$
I.e. $(f(b_i))$ converges to the same limit $L$.

So the function $g$ is sequentially continuous at $a \in X$ and therefore continuous which defines $L$ as the limit of $f$ at $a$.