# Continuous mapping between topological spaces

Let $X$ and $Y$ be topological spaces. The mapping $f:X\to Y$ is continuous if the preimage of the open set is an open set.

"If the topology on $X$ is finer it is "easier" for $f$ to be continuous" ($*$)

Question: As far as I know, the topological space consists only of open sets. For my imagination, in any case , topology being finer or coarser (it's elements are always open) if the preimage of the open set is open then $f$ is always continuous.

I can not imagine how can we say "easier", I mean, how can we compare? Are there any other cases when the topology consists not only of open sets, that make $f$ not "easy" to be continuous? If yes, how can it be topology?

I am really confused now.

I appreciate any help.

## 2 Answers

On the same set it is possible to have many different topologies. For example on $\mathbb R$ you have not only the usual topology, but also the discrete topology (all subsets are open) as the finest and the indescrete (only $\emptyset$ and $\mathbb R$ are open) as the coarsets possible topology - and many more. Now if we fix a map $f\colon X\to Y$ and fix a topology on $Y$, but consider different topologies on $X$, then the continuity of $f$ depends on the topology on $X$. For example, with the discrete topology on $X$, all such maps $f$ are continuous; with the indiscrete it is much less "likely" that a given $f$ is continuous; with other topologies, the situation is inbetween: If $f$ is continuous wrt. to a coarser topology (e.g. wrt. to the indiscrete topology) the it is also continuous wrt. to this topology, whereas the converse need not hold.

Well, topological space consists of points - basically, it's a set. Some subsets of this set are distinguished as "open sets". The finer the topology, the more subsets are considered "open". Consequently, if we consider mapping $f\colon X\to Y$, if it's continuous for some topology on $X$, it's also continuous for any finer topology (since preimages of open sets in $Y$ are open in the coarser topology on $X$, they're open in finer as well). The converse does not hold.