1. ## Uniformly continuous

Wazzup!

How do I get around proving (by an indirect argument) that a continuos mapping f: X -> Y between metric-spaces (X,d_X) and (Y,d_y) is uniformly continuous in every sequentially compact subset K of X.

2. My suggestion is to suppose that $f$ is not uniformly continuous. Then, there exists $\varepsilon> 0$ such that for every $\delta>0$ there exist $x,y\in X$ such that $d(x,y)<\delta$ but $d(f(x),f(y))\geq \varepsilon$.

Try to construct a sequence that behaves badly. Keep in mind that the space is sequentially compact.

3. How do I use tha fact that the space is sequentially compact.

4. You first need to construct some kind of sequence out of the failure of uniform continuity. For example, for $\delta=1$ we can find points $x_1,y_1\in X$ such that $d(x_1,y_1)<1$ but $d(f(x_1),f(y_1))\geq \varepsilon$. For $\delta=1/2$, we can find $x_2,y_2\in X$ such that $d(x_2,y_2)<1/2$ but $d(f(x_2),f(y_2))\geq \varepsilon$. Continuing the process, we can find $\delta=1/n$ such that $d(x_n,y_n)<1/n$ but $d(f(x_n),f(y_n))\geq \varepsilon$.

This gives you two sequences $\{x_n\}$ and $\{y_n\}$ such that $d(x_n,y_n)<1/n$ and $d(f(x_n),f(y_n))\geq \varepsilon$ for every $n$. Use sequential compactness here to relate them to the continuity of $f$.

5. Originally Posted by mus
Wazzup!

How do I get around proving (by an indirect argument) that a continuos mapping f: X -> Y between metric-spaces (X,d_X) and (Y,d_y) is uniformly continuous in every sequentially compact subset K of X.
Note: This is more high-powered for than roninpro's suggestion. If it's gobbeldygook to you just ignore it.

It's trivial that for metric spaces sequential compactness and compactness are synonymous. So, let $\varepsilon>0$ be arbitrary and cover $f(K)$ with the open balls $\left\{B_{\varepsilon}(f(x))\right\}_{x\in K}$. Clearly then we have that $\left\{f^{-1}\left(B_{\varepsilon}(f(x))\right)\right\}_{x\in K}$ is an open cover for $K$. Then, since $K$ is compact we know that the open cover $\left\{f^{-1}\left(B_{\varepsilon}\left(f(x)\right)\right)\ri ght\}_{x\in K}$ admits a Lebesgue number $\delta$. Note then that if $d_X(x,y)<\delta$ then we have that $\text{diam}\{x,y\}<\delta$ (this last part was pedantic, but just in case) and so by definition we have there is some $f^{-1}\left(B_{\varepsilon}(f(x_0))\right)$ such that $x,y\in f^{-1}\left(B_{\varepsilon}(f(x_0))\right)$ and so $f(x),f(y)\in B_{\varepsilon}(f(x))$ and thus $d_Y(f(x),f(y))<2\varepsilon$.