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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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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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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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