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5

I think that your proof is correct. However why to proceed by contradiction? Consider an open cover. One of the open set $\mathcal O$ contains $0$. As $0 \in \mathcal O$, there is an open disk centered on $0$ included in $\mathcal O$. This disk contains all but a finite number of the points $1/n$. Consider the finite open sets of the cover that cover the ...

3

a) OK with your proof. b) Yes. A product of two (non empty) connected sets is a connected set and the image of a connected set by a continuous map is connected. c) Yes. By triangular inequality. d) Yes. The image of a compact set by a continuous map is compact and the product of two compact sets is compact.

3

Something basic is First Concepts of Topology, a positively reviewed book that is both cheap and short, and it also reads like a narrative (there are exercises if you want to enhance your understanding, but you obviously do not have to do them unless you want--full solutions to the exercises are available in the back of the book). Perhaps more along the ...

1

Let $\tau$ be the collection of order-closed subsets of $X$, we show that this defines a topology on $X$. We have that $\emptyset \in \tau$ vaculously and $X \in \tau$ trivially. $\tau$ is closed under arbitrary intersection, this can easily be shown. We show that $\tau$ is closed under finite union. Let $B_{1}, B_{2} \in \tau$. We want to show that ...

1

Let $x\in\mathbb Q\cap [0,1)$. Suppose there exists $r>0$ with $B_r(x)\subseteq\mathbb Q\cap [0,1)$. Then $y=x+r/2\in B_r(x)$. If $y$ is irrational, this is a contradiction. If $y$ is rational, there exists an irrational $z$ such that $x<z<y$ and $z\in B_r(x)$. This is also a contradiction, so $r=0$ is the only number such that ...

1

I agree that $F$ is bounded and countable. However $F$ is not closed. $\{0\}$ is a limit point that is not in $F$. $F \cup \{0\}$ is a compact set.

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