Is the following condition implied by "proper map"? As a part of a text book I'm reading, we were talking about a proper map $f:\:X\to Y$, for $X,Y$ topological spaces with "good qualities" (we can assume pretty much anything). We had a sequence $x_i \in X$ such that $f(x_i)\to y\in K$ for $K\subset Y$ compact.
It was mentioned that $x_i$ must have a converging sub-sequence. I didn't understand why, as it was not explained, but it feels like it is stronger than $f$ being proper.
if we new that $\forall i>i_0\: f(x_i)\in K$ then we could say that $(x_i)_{i>i_0} \subset f^{-1}(K)$, the latter being compact, and from properness get the desired result (assuming that compactness implies sequencial compactness, which is ok), but the way I see it we can have $\forall i\in\mathbb N\:f(x_i)\not\in K$ and get away with that.
 A: My definition of proper map is (you don't state yours) for $f:X \rightarrow Y$ is that $f$ is continuous, closed, onto and $f^{-1}[\{y\}]$ compact for every $y \in Y$. It implies the fact (often taken as the definition) that $f^{-1}[K]$ is compact for any compact $K \subseteq Y$.
Then if $X$ is a sequential space (sequentially closed sets are closed) the statement indeed holds: the sequence $(x_n)$ lives in  the compact set $f^{-1}[K]$, where $K: =\{f(x_n) : n \in \mathbb{N}\} \cup \{y\}$ which is compact in any space $Y$ (as a convergent sequence including its limit).
In a nice space like a sequential space (this includes all first-countable, hence all metric spaces), $f^{-1}[K]$ is then sequentially compact and the subsequence must exist. So only $X$ has to be (moderately) "nice", for this fact to hold.
A: Assume that $X,Y$ behave like locally compact metric spaces. Since $f$ is proper, the inverse image of a compact subset is compact. The point $y$ has a compact neighborhood $V$, there exists $N$ such that $n>N$ implies that $f(y_n)\in V$. $f^{-1}(V)=W$ is compact since $f$ is proper and $x_n\in V, v>N$. Since $W$ is compact we can extract a converging subsequence from $(x_n)$.
