Question: Let $\{G\_i\}\_{i \in \Lambda}$ be a family of compact sets in a Hausdorff topological space. Show that the intersection $\bigcap\_{i \in \Lambda} G\_i$ is compact.

My attempt was:

Since $G\_j$ for some $j \in \Lambda$ is compact, there exists an open, finite covering. And $\bigcap\_{i \in \Lambda} G\_i \subset G\_j$. So I can take the same covering for the intersection as for $G\_j.$

Is that correct?

Answer: 
**No.** If we consider an infinitely differentiable function $f(x)$, then we can write down it's Taylor series about the point $x = a$ as
$$\sum\_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n.$$ Now, if we consider the translation $h(x)=f(x+a)$ and compute its Maclaurin series, we get $$\sum\_{n=0}^{\infty}\frac{h^{(n)}(0)}{n!}x^n = \sum\_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}x^n.$$ Note that $$\sum\_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n \ne \sum\_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}x^n$$ because they clearly disagree at $x = a$. What we *can* say, however, is that the Taylor series for $f(x)$ is the Maclaurin series for $g(x) = f(x+a)$ **shifted to the right by $a$.**
Let $ f $ be a function defined arround $ x= a $, let us say at $ (a-\eta,a+\eta)$.

then, the function $ g $ given by $ g(x)=f(x+a) $ will be well defined at $(-\eta,\eta) $.

If $$ g(x)=P\_n(x)+x^n\epsilon\_1(x)$$

with $\lim\_{x\to 0}\epsilon\_1(x)=0$

then

$$f(x)=g(x-a)=$$
$$P\_n(x-a)+(x-a)^n\epsilon\_2(x)$$
with $ \lim\_{x\to a}\epsilon\_2(x)=0$.