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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.

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  • $\begingroup$ I think you want the union in the definition of closure to be an intersection $\endgroup$ – Zach Effman May 5 '15 at 1:41
  • $\begingroup$ @ZachEffman I edited. Thanks! $\endgroup$ – user219081 May 5 '15 at 1:45
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Hint: Is it true when $f$ is an inclusion? Consider a few examples.

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  • $\begingroup$ I am not sure what an inclusion is $\endgroup$ – user219081 May 5 '15 at 12:17
  • $\begingroup$ @AlyssaWallace An inclusion is when $X$ is a subspace of $Y$. For example, the $x$-axis includes into the $xy$-plane. $\endgroup$ – Slade May 5 '15 at 16:59
  • $\begingroup$ Is this a false statement without inclusion? $\endgroup$ – user219081 May 6 '15 at 13:45
  • $\begingroup$ @AlyssaWallace Let me be clear: look at a few inclusions. Keep looking and you will find a counterexample. $\endgroup$ – Slade May 6 '15 at 16:38
  • $\begingroup$ after trying to look back at my book... I have no examples of continuous functions for this topic. My only examples are piece-wise functions. Any suggestions of where to look? $\endgroup$ – user219081 May 6 '15 at 18:44

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