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Let $A \subset \mathbb{R}$ be a connected set and let $f:A\to \mathbb{R}$ be a continuous function. We define $f(A)$ to the image of $f$. Show that $f(A)$ is connected.

I thought about maybe using contradiction. So that we assume $\exists X,Y \in \mathbb{R}$ such that $X\cup Y =f(A)$ and $X\cap Y = \emptyset$ and use this to contradict that $f$ is continuous. I am stuck here and some hints to get me in the right direction would be much appreciated. Thanks!

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Consider the sets $f^{-1}(X)$ and $f^{-1}(Y)$. –  David Mitra Feb 3 '13 at 16:22
    
If $O$ is open in $\mathbb R$, then $f^{-1}(O)$ is open in $A$ (because $f$ is continuous). –  Stefan Feb 3 '13 at 16:23

2 Answers 2

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Using @gnometorule's hint, recall that $$f^{-1}(U\cup V)=f^{-1}(U)\cup f^{-1}(V),$$ where $$f^{-1}(U)=\left\{x\in X:\, f(x)\in U\right\}.$$ Now, if $X$ is connected, suppose that $f(X)$ is not connected, where $f$ is continuous. Then there exists a separation of $f(X)$ into disjoint open sets, suppose $f(X)= U\cup V$. Then $$X=f^{-1}(U\cup V)=f^{-1}(U)\cup f^{-1}(V).$$ Since $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint, this shows $X$ can be separated into disjoint open sets, a contradiction since we're given $X$ is connected. Thus, $f(X)$ must be connected.

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Thanks for your answer! –  Slugger Feb 3 '13 at 16:36
    
You're welcome. I also responded to your other comment. –  Clayton Feb 3 '13 at 16:38

If you use $f^{-1}(U \cup V) = f^{-1}(U) \cup f^{-1}(V)$, which is always true, your idea will show a contradiction in the next step. Do you see it?

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I thought about using this but how can we be sure that $f^{-1}$ exists. It also doesn't assert that $f$ is one-to-one. –  Slugger Feb 3 '13 at 16:25
    
@TeunVerstraaten $f^{-1}(U)$ here is the preimage of a set $U$, not the inverse function. –  Ayman Hourieh Feb 3 '13 at 16:27
    
So $f^{-1}(X \cup Y) = f^{-1}(X) \cup f^{-1}(Y)=A$, therefore implying that $A$ is not connected and thus reaching a contradiction. Thanks! do we need to check that $f^{-1}(X) \cap f^{-1}(Y)=\emptyset$? –  Slugger Feb 3 '13 at 16:34
    
@TeunVerstraaten: Note that if the intersection is nonempty, then $X\cap Y$ is nonempty, contradicting the assumption. –  Clayton Feb 3 '13 at 16:38

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