# Let $f:(X, \mathfrak T_X) \rightarrow (Y, \mathfrak T_Y)$ be a continuous function. Then $f(Cl(A) = Cl(f(A))$.

Let $f:(X, \mathfrak T_X) \rightarrow (Y, \mathfrak T_Y)$ be a continuous function. Then $f(Cl(A) = Cl(f(A))$.

My definition of closure is:
Let $(X,\mathfrak T)$ be a topological space and let $A \subseteq X$ . The closure of $A$ is $Cl(A) = \bigcap \{U \subseteq X: U$ is a closed set and $A \subseteq U\}$ Based on this I know $A \subseteq Cl(A)$

My definition of continuous is "A function $f : \mathbb R \rightarrow \mathbb R$ is said to be continous if for each open subset $V$ of $\mathbb R, f^{-1}(V)$ is an open subset of $\mathbb R$.

I am supposed to determine if this is true or false and if true prove it and if false give a counterexample. I have the definitions but I really do not even know where to start. I have all the relevant definitions gathered but do not know where to go from here.

• I think you want the union in the definition of closure to be an intersection – Zach Effman May 5 '15 at 1:41
• @ZachEffman I edited. Thanks! – user219081 May 5 '15 at 1:45

Hint: Is it true when $f$ is an inclusion? Consider a few examples.
• @AlyssaWallace An inclusion is when $X$ is a subspace of $Y$. For example, the $x$-axis includes into the $xy$-plane. – Slade May 5 '15 at 16:59