# metric spaces

• May 2nd 2009, 05:39 AM
jin_nzzang
metric spaces
Let (X,d) and (Y,p) be metric spaces, and assume Y is complete. Let f: X→Y and x∈X.
Show that f has a limit at x if and only if given any ε>0 there exists δ>0 such that whenever y, z ∈B_δ(x)＼｛x｝,p(f(y),f(z)) < ε.

• May 3rd 2009, 01:33 AM
Opalg
Quote:

Originally Posted by jin_nzzang
Let (X,d) and (Y,p) be metric spaces, and assume Y is complete. Let f: X→Y and x∈X.
Show that f has a limit at x if and only if given any ε>0 there exists δ>0 such that whenever y, z ∈B_δ(x)＼｛x｝,p(f(y),f(z)) < ε.

Let $(x_n)$ be a sequence in X that converges to x. Then $(x_n)$ is a Cauchy sequence in X. Use the given ε-δ condition to conclude that $(f(x_n))$ is a Cauchy sequence in Y. The completeness of Y tells you that $(f(x_n))$ converges to some element w in Y.

It remains to show that for every sequence that converges to X, its image under f converges to the same limit in Y. So suppose that $x_n\to x$ with $f(x_n)\to w$, as above, and also that $y_n\to x$ with $f(y_n)\to v$. Then $d(x_n,y_n)\to0$, and therefore $p(f(x_n),f(y_n))\to0$, from which it follows that $p(w,v)=0$ and hence v=w.

Thus for every sequence that converges to x, its image under f converges to w. Therefore $\textstyle\lim_{y\to x}f(y)$ exists and is equal to w.