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Define $\phi : C[-1,1] \to \mathbb R$ by $\phi(f) = f(0)$. Note that $|\phi(f)-\phi(g)| \leq d_\infty(f,g)$ for every $f$ and $g$ in $C[-1,1]$. Thus, $\phi$ is continuous and then $A = \phi^{-1}([2,4])$ is closed in $(C[-1,1],d_\infty)$.
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Topology is a rather broad field, so there are very many interesting spaces to cover.
My personal favorite is the Sierpinski-space, which is the space $\{\bullet, \circ\}$, with precisely one nontrivial open set $\{\circ\}$. It is interesting out of the following reasons:
It is the smallest nontrivial topological space
It can be used to detect or classify ...
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Let $p : C^{n+1} \setminus \{0\} \to \mathbb CP^n$ denote the quotient map.
The sets $V_i = \{ z \in \mathbb C^{n+1} \mid z_i \ne 0 \}$ ares open in $\mathbb C^{n+1}$, thus open in $\mathbb C^{n+1} \setminus \{0\}$ since $0 \notin V_i$. Clearly $\bigcup_{i=0}^n V_i = \mathbb C^{n+1} \setminus \{0\}$. We have $U_i = p(V_i)$, thus the $U_i$ cover $\mathbb CP^n$...
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I think you put your finger of the issue. $A$ and $B$ don't have to be compact.
Let $$A= \bigcup_{n=1}^\infty\left[2n+1,2n+2-\frac1n\right]\\
B= \bigcup_{n=1}^\infty\left[2n,2n+1-\frac1n\right]
$$ and left $f$ be identically $0$ on $A$ and identically $1$ on $B$. There are points in $A$ and $B$ arbitrarily close to one another.
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Define $i : D^n \to S^n, i(x) = (x,\sqrt{1-\lvert x \rVert^2})$. This map embeds $D^n$ as the closed upper hemisphere of $S^n$.
You know that $\mathbb P^n$ is the quotient space of $S^n$ obtained by identifying antipodal points. Let $p : S^n \to \mathbb P^n$ denote the quotient map. The map $H = p \circ i : D^n \to \mathbb P^n$ has the property $H(x) = H(y)$ ...
1
Let $A = \{f \in C[-1, 1] : 2 \leq f(0) \leq 4 \}$. We'll show this is closed by showing that the complement is open. Suppose $f \in A^{c}$. Then $f(0) > 4$ or $f(0) < 2$. In either case, there is some $\epsilon > 0$ such that $(f(0) - \epsilon , f(0) + \epsilon) \cap [2 , 4] = \emptyset$. This should be intuitive, but rigorously, this follows ...
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Let $(f_n)_{n\in \Bbb N}$ be a convergent sequence of members of $A,$ converging to $g.$ For any $n\in \Bbb N$ we have $|f_n(0)-g(0)|\le d(f_n,g).$ And $d(f_n,g)\to 0,$ so $f_n(0)-g(0)\to 0.$ So $g(0)=\lim_{n\to\infty}f_n(0)\in [2,4]$ because every $f_n(0)\in [2,4]$ and $[2,4]$ is closed in $\Bbb R.$ So $g\in A.$
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First note that a function has the property you are looking for iff it is everywhere surjective:
everywhere surj $\rightarrow$ preimage dense: Given $x\in \mathbb{R}$, $f((x-\varepsilon,x+\varepsilon))=\mathbb{R}$, so $\forall_y f^{-1}(y)\in (x-\varepsilon,x+\varepsilon)$. Since this must hold for every $\varepsilon>0$, we get the claim .
preimage dense $...
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Here is a counterexample, which highlights the fact that completeness is not independent of the metric.
Take $X = \mathbb R$, which is a 1-dimensional topological vector space. Use the homeomorphism $f : \mathbb R \to (-\pi/2,\pi/2) $ given by $f(x) = \tan^{-1}(x)$ to transport the metric from $(-\pi/2,\pi/2)$ back to $\mathbb R$, obtaining $d(x,y) = \left|f(...
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Suppose that $T$ is the graph of a function, that is, suppose that there exists $g : X \to Y$ such that $T = \{(x,y) \in X \times Y : y = g(x)\}$. Now, take $a \in X$ and observe that $T \cap (\{a\} \times Y)$ is non-empty since $(a,g(a)) \in T \cap (\{a\} \times Y)$. We claim that $T \cap (\{a\} \times Y) = \{(a,g(a))\}$. Indeed, if $(x,y) \in T \cap (\{a\} ...
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For the first part of the proof the fact that $f[X]$ is open in $\Bbb Z$ is irrelevant: all that matters is that the only non-empty, connected subsets of $\Bbb Z$ are the singletons.
The other direction is fine, though it can be done more simply without the pasting lemma. Just prove the contrapositive. If $X$ is not connected, there is a continuous function $...
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