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Please help me do the following.

Suppose that $f:\mathbb{R}^m\to\mathbb{R}$ satisfies two conditions:

(i) For each compact set $K$, $f(K)$ is compact.

(ii) For any nested decreasing sequence of compacts $(K_n)$, $$f\left(\bigcap K_n\right)=\bigcap f(K_n).$$ Prove that $f$ is continuous.

Property (ii) implies the following: If $(x_i)$ is a sequence in $\mathbb{R}^m$ converging to $x$ such that $f(x_1)=f(x_2)=\ldots$, then $f(x)=f(x_1)$. I think they are in fact equivalent, but I'm not sure.

Edit: Sorry for the confusion, the second condition is now corrected.

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Excuse me, english is not my native language. What is a nested decreasing sequence of compact sets? – matgaio Apr 23 '12 at 7:31
Are you sure this is a union in (ii)? – user20266 Apr 23 '12 at 7:37
@matgaio Each $K_n$ is compact and $K_{n+1}\subset K_n$. On another note, the union in (ii) should probably be an intersection. Indeed,if one of $K_n=\varnothing$, then the LHS is empty, whereas the RHS is nonempty if at least one of the $K_n$ is nonempty. – Benji Apr 23 '12 at 7:41
I think you mean $f(\cap K_n) = \cap f(K_n)$ in ii. – Kuku Apr 23 '12 at 7:43
I thought the same about the condition ii), in order to get some convergence property to this function. – matgaio Apr 23 '12 at 7:45

Since the question was edited, my answer also requires an update.

We show that $f$ is continuous at an arbitrary $x\in\mathbb{R}^{n}$. Let $U\subset \mathbb{R}$ be an open neighborhood of $f(x)$. Note that property $(ii)$ implies that $$\{f(x)\}=f\Big(\bigcap_{n=1}^{\infty} \bar{B}(x,\frac{1}{n})\Big)\overset{(ii)}{=}\bigcap_{n=1}^{\infty}f\big(\bar{B}(x,\frac{1}{n})\big),$$ where $\bar{B}(x,\frac{1}{n}):=\{y\in\mathbb{R}^{n}:\|x-y\|\leq \frac{1}{n}\}$ is the closed $\frac{1}{n}$-radius ball around $x$. By property $(i)$ each $f\big(\bar{B}(x,\frac{1}{n})\big)$ is compact (since by Heine-Borel each $\bar{B}(x,\frac{1}{n})$ is), and $\{f(x)\}$ is thus the intersection of nested non-empty compact sets in a Hausdorff space. Since $U$ is the neighborhood of this intersection, there exists $k\in\mathbb{N}$ so that every member of the intersection is inside $U$ starting from $k$, i.e. $f\big(\bar{B}(x,\frac{1}{n})\big)\subset U$ for all $n\geq k$. This property is very straight-forward to prove, it is done for example here: Nested Sequence of Non-Empty Compact Sets

Now choose e.g. $\delta=\frac{1}{k+1}>0$, whence $f(B(x,\delta))\subset U$. We have shown that $f$ is continuous at $x$, whence $f$ is in fact a continuous function (since the choice of $x$ was arbitrary).

Old answer (before OP edited the question) It seems that every sequence would have that property that you mentioned.. The property $(ii)$ makes this kind of weird and the property $(i)$ plays no role at all. Someone please correct me if I overlooked something.

Let $x\in\mathbb{R}^{m}$ be arbitrary and $(x_{n})_{n=1}^{\infty}\subset \mathbb{R}^{m}$ converging to $x$. The sets $K_{i}:=\{x_{n}:n\geq i\}\cup \{x\}$ are compact for each $i\in\mathbb{N}$ as closed and bounded subsets of $\mathbb{R}^{m}$ by Heine-Borel. Moreover, $(K_{i})_{i=1}^{\infty}$ is a decreasing sequence of compact sets with $\cap_{i=1}^{\infty} K_{i}=\{x\}$. By property $(ii)$ we have $f(\{x\})=f(\cap_{i=1}^{\infty} K_{i})=\cup_{i=1}^{\infty}f(K_{i})$, so infact $f(K_{i})=\{f(x)\}$ for all $i\in\mathbb{N}$ and in particular the sequence $(f(x_{n}))_{n=1}^{\infty}$ is a constant sequence of $f(x)$ converging to $f(x)$. Hence $\lim_{n\to\infty}f(x_{n})=f(x)$, which shows that $f$ is continuous at $x$. Moreover, since the choice of $x$ was arbitrary, we have that $f$ is a continuous function.

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In fact more can be said if condition (ii) is correct as stated: $f $ must be constant. – Benji Apr 23 '12 at 8:16
Yeah, that's correct. – T. Eskin Apr 23 '12 at 8:23
It seems the OP has changed the question, and so your answer no longer holds. – M Turgeon May 26 '12 at 21:35
Thanks @MTurgeon, I updated my answer:) – T. Eskin May 29 '12 at 7:31

Let's write $B_r(a)=\{x\in\mathbb R^n\colon |x-a|\le r\}$. Here is a lemma that you should be able to prove yourself.

Lemma. If $f$ is bounded in a neighborhood of $a$ and $\bigcap_{n=1}^{\infty}\overline{f(B_{1/n}(a))}=\{f(a)\}$, then $f$ is continuous at $a$.

(Idea of the proof of the lemma: if $f$ is not continuous, there is a sequence $x_n\to a$ such that $f(x_n)\to y\ne f(a)$. Show that $y$ is contained in the intersection.)

With the help of the lemma the proof proceeds as follows. Let $a$ be any point. Since $f(B_1(a))$ is compact, the function $f$ is bounded in a neighborhood of $a$. By the assumption (i), $f(B_{1/n}(a))$ is closed for every $n$. Thus, $$\bigcap_{n=1}^{\infty}\overline{f(B_{1/n}(a))}=\bigcap_{n=1}^{\infty}f(B_{1/n}(a)) = \{f(a)\}$$ where the last step is based on (ii).

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