Every topological space $X$ has the initial topology with respect to the family of continuous functions from $X$ to the Sierpiński space. I am currently reading about initial topologies w.r.t. the Sierpiński space, and on Wikipedia I read the following

Every topological space $X$ has the initial topology with respect to the family of continuous functions from $X$ to the Sierpiński space.

Could someone explain to me what that means, is it that every topology could be represented as an initial topology w.r.t. to Sierpiński space?
 A: Consider a topological space $X$ (let's call its topology $\mathcal{T}$). Let $S$ be the Sierpiński space. Let
$$F=\{\text{continuous functions }f\colon X\to S\}.$$
Let $\mathcal{T}'$ be the initial topology on $X$ for the family of functions $F$ -  that is, the smallest topology on $X$ under which every function in $F$ is continuous (this is the intersection of all topologies on $X$ that have this property). Then the statement is saying that in fact $\mathcal{T}=\mathcal{T}'$.
A: If $X$ is a topological space, and $S$ is the Sierpinski Space, then the set of continuous functions $f:X\to S$ is in $1-1$ correspondence with the open sets of $X$. So the topology is entirely determined by the set of continuous functions from $X$ to $S$.
Let $S=\{0,1\}$ be the space with open sets $\emptyset,\{0\},\{0,1\}$. Then for $U\subseteq X$ open, we define $f_U:X\to S$ by $f_U(x)=0$ if $x\in U$ and $f_U(x)=1$ if $x\not\in U$. This is continuous. 
We can also see that if $f:X\to S$ is continuous, then $f=f_U$ where $U=f^{-1}(\{0\})$.
A: For every space you can do the following. Take any space $Y$ and let $\mathcal F$ be the family of all continuous maps from $X$ to $Y$. Now equip $X$ with the initial topology with respect to this family. It is the coarsest topology making all $f\in\mathcal F$ continuous and has as subbase the preimages $f^{-1}(U)$ of open sets $U$ in $Y$ under these $f$'s. Clearly, this topology will always be coarser than the topology you started with, since by construction the original topology made these maps continuous.
Now let $\mathcal F$ be the family of all continuous functions from $X$ to the Siepiński space $S=\{a,b\}$, where $\{a\}$ is the only non-trivial open set. Then the statement says that the initial topology w.r.t. $\mathcal F$ will not be strictly coarser than the $\tau$. To show this, all you have to do is to take an open set $U$ and find a continuous map $f:X\to S$, such that $U=f^{-1}(\{a\})$.
One can try other spaces for $Y$. For instance, take $Y=\Bbb R$ or $Y=[0,1]$. Then you get the spaces with a well-known property. Can you find out which it is?
