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hornejada1c 2022-07-15 Answered
Let and $f:X\to \mathbb{R}$ defined by $f\left(x,y\right)=x$ . I want to find $C\subset X$ closed such that $f\left(C\right)\subset \mathbb{R}$ isn't closed. How to prove that f is a closed map?
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## Answers (1)

Kaylie Mcdonald
Answered 2022-07-16 Author has 19 answers
Step 1
f i understand correctly, $X=\left\{\left(x,0\right):x<0\right\}\cup \left\{\left(x,y\right):x\ge 0\right\}$ . So, if we take $C=\left\{\left(x,0\right):x<0\right\}\cup \left\{\left(0,y\right):y\in \mathbb{R}\right\}$ then C is closed because $X-C$ is open in ${\mathbb{R}}^{2}$ but when we apply f (which is the projection map actually, i.e. continuous 1-1 onto), we have $f\left(C\right)=\left(-\infty ,0\right)$ which is obviously open in $\mathbb{R}$ .

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