# Tag Info

8

Not necessarily; consider $f:(-2\pi,2\pi)\to[-1,1]$ given by $f(x)=\sin(x)$.

5

Compactness is quite possibly the most important topological property, both in topology itself, and in other fields which build on topological results. The reason is that compactness allows us to attribute properties of finite sets to infinite sets. This comes from the definition: any open cover has a finite subcover. In very broad terms, a compact space ...

4

HINT: Show that for every function $F\colon X\to X$, and every $A\subseteq X$: If $F(A)\subseteq A$, then $F(F(A))\subseteq F(A)$.

3

Hint: Try finding a sequence in $\mathbb{Q}\cap[0,1]$ that does not have a convergent subsequence in $\mathbb{Q}\cap[0,1]$.

3

The step in question is why does it follow that \begin{align*} \left | \int_0^x (x-t) f(t) dt - \int_0^x (x - t) g(t) dt \right | \leq d_\infty(f,g) \int_0^x (x-t) dt. \end{align*} This follows easily from the defintion of $d_\infty(f,g) = \sup_{s \in [0,1]}|f(s) - g(s)|$ and the the estimate \begin{align*} \left | \int_0^x (x-t) f(t) dt - \int_0^x (x - t) ...

3

We have the inequality $$\left\lvert \int_a^b f \right\rvert \leqslant \int_a^b \lvert f \rvert,$$ basically as a generalisation of the triangle inequality. Applying this to your integral gives $$\left\lvert \int_0^x (x-t) (f(t)-g(t)) \, dt \right\rvert \leqslant \int_0^x (x-t) \lvert f(t)-g(t) \rvert \, dt.$$ If $u$ is a nonnegative function on $(a,b)$ ...

3

Choose $X = \mathbb{R}$ and let $K$ to be the Cantor set. Then $X \setminus K$ has infinitely many connected components.

3

You could express it a bit better, but the argument is basically correct. Suppose that $C$ is a compact subset of $\Bbb Q$ with $0$ that contains an open nbhd of $0$; then there is an $\epsilon>0$ such that $(-\epsilon,\epsilon)\cap\Bbb Q\subseteq C$. Let $f:\Bbb Q\to\Bbb R$ be the identity map; $f$ is continuous, so $f[C]$ is compact, and ...

3

Fix $g\in L^1$ non-negative and consider $$K:= \{f \in L^1 : |f| \leq g\}$$ Then $K$ is bounded by $\|g\|_1$ and uniformly integrable (by the absolute continuity of $\int g$), and so it is weakly compact by the Dunford-Pettis theorem. (See this) Edit: As pointed out by Norbert, it remains to show that $K$ is weakly closed. Since it is convex, it suffices ...

2

Take any finite cover of $X$ in $\mathbb{R}^n$ and take the intersections of those sets with $X$. Then those will be open in $X$ (in the subspace topology - this is probably how he means you to interpret this) and their union will be $X$.

2

For simplicity of the construction, consider $(-1,1)$. Let $K = \{0\} \cup \{\frac1{n+1}: n \in \Bbb N\}$. Define $K_n$ as: $$K_n = \left[-\frac{n}{n+1},\frac{n}{n+1}\right] \setminus \bigcup_{k = n}^\infty \left(\frac1{4k+3},\frac1{4k+1}\right)$$ Then it is manifest that $K_n$ is increasing and its union is $(-1,1)$. However, for each $n$, $\dfrac1{4n+2} ... 2 There are no non-trivial compact subrgroups of the group$G$, because each non-unit element$g\in G$generates an unbounded subgroup$\langle g\rangle=\{g^n:n\in\Bbb Z\}$which cannot be contained in a compact subset of the space$G$. 2 If$f$has no infimum$> 0$, then$f^{-1}((\epsilon, \infty)) \ne M$for any$\epsilon > 0$. 2 Let$y_n \in \bar Y$. Then by definition of$\bar Y$, there are$x_n \in Y$so that$d(x_n , y_n) < \frac 1n$. Then as$Y$is precompact, there is$x\in X$so that$x_n \to x$. Thus$x\in \bar Y$. Also$y_n \to x$. Thus$\bar Y$is compact. 2 Hint. Let$L_i = \bigcap_{j=1}^i K_j$, then the$K_1 \setminus L_i$form an open cover of$K_1$. Now use$K_1$'s compactness to find an$i$such that$K_1 = K_1 \setminus L_i$(note that the$K_1 \setminus L_i$are an increasing sequence), hence$L_i = \emptyset$. 2 No. Take$X = [-1,0)\cup (0,1]$. Then$d(-1,1) = 2 = \sup_{u,v\in X} d(u,v)$but X is clearly non compact (for instance the open cover$ ([-1,1]\setminus[-\frac1n,\frac1n])_{n\in \mathbb N}$doesn't have a finite subcover). 2 You have $$\left|\int_0^x (x - t) (f(t) - g(t))\ dt \right|.$$ I assume your notation$d_\infty(f, g)$stands for the$L^\infty$norm$\|f - g\|_\infty$. In this case I can shed some light on the next step. Assume for convenience that$f$and$g$are continuous (the next step holds regardless, I just think it's more intuitive if they are). Then we can crash ... 2 Succinctly, we have for$x\in I=[0,1]\begin{align} \left|\int_0^x (x-t)\left(f(t)-g(t)\right)\,dt\right|&\le\int_0^x (x-t)\left|f(t)-g(t)\right|\,dt \tag 1\\\\ &\le \sup_{t\in [0,x]}\left|f(t)-g(t)\right|\int_0^x (x-t)\,dt \tag 2\\\\ &\le\frac12\,\sup_{t\in [0,x]}\left|f(t)-g(t)\right| \tag 3 \end{align} sincef$and$g$(and hence ... 2$\mathbb{Q}\cap[0,1]$is dense in$[0,1]$2 First: consider a function$f:X\rightarrow Y$, and$A\subset B \subset X$. I shall show that$f(A)\subset f(B)$. Consider an element$x\in f(A)$, and an element$y\in A$such that$f(y)=x$. As$A\subseteq B$and$y\in A$, we have$y\in B$. Therefore,$f(A)\subseteq f(B)$. So if$X_i\subseteq X_{i-1}$, then$X_{i+1}=f(X_i)\subseteq f(X_{i-1})=X_i$. As ... 2 In$\mathbb{R}$, let$X = \{0\} \bigcup \{1, \frac{1}{2}, \frac{1}{3}, \ldots \}$. Then$X$is compact and has infinitely many isolated points. 2 The set$\{x_n\}$is in general not closed, but its closure$\overline{\{x_n\}}$is. You just need to prove that the closure of a bounded set is again bounded to correct your proof. 1 This does not work without the assumption that$X$is normal. For a counterexample, consider the $$**\textbf{sequence with two limits}**$$ Let$X=\{0,0'\}\cup\{1/n\mid n\in\Bbb N\}$, where$\{1/n\mid n\in\Bbb N\}$is discrete and a neighborhood of$0$or$0'$is a set containing almost all of the points$1/n$. This space is compact but not normal, and ... 1 Again, this is false. Take $$C = \prod_{i=1}^{n} [0,1]_i \subset \mathbb{R}^n.$$ Then,$C$is a perfect set in$\mathbb{R}^n$, so every point of$C$is a limit point. Ergo, there does not exist a point such that it can be surrounded by a neighborhood not containing another point of$C$. In general, any compact connected subset of$\mathbb{R}^n$will not ... 1 Clearly$\mathscr{A}$must contain all finite subsets of$X$and be closed under finite unions. It must also be closed under subsets, since if$A\subseteq B$, and$\operatorname{cl}B$is compact, then$\operatorname{cl}A$is compact as well. In particular, if$X\in\mathscr{A}$, then$\mathscr{A}=\wp(X)$. All of this makes$\mathscr{A}$an ideal in$\wp(X)$, ... 1 Hint: Find a sequence of closed, non-empty sets$C_n\subseteq [0, 1]\cap \Bbb Q$for$n \in \Bbb N$such that$C_n \subseteq C_{n-1}$and$\bigcap_{n = 1}^\infty C_n = \emptyset$. 1 You can cover it with infinite number of open sets such that no open set overlaps (open intervals with irrational endpoints). If$\bigcup U_j = X$, but$U_j\cap U_k = \emptyset$if$k\ne j$you can't reduce the sets to a finite covering. 1 You said, "A closed subspace of a compact space is compact." and then you tried using$[0,1]$as the compact space. But$[0,1]$is compact as a subset of the reals, as a subset of$\mathbb{Q}$, it's what you are trying to understand. The countable covering of$[0,1]$in$\mathbb{Q}$by point sets in$[0,1] \cap \mathbb{Q}$has no finite subcover in ... 1 Hint: let$X_0 = X$and$X_{i+1} = T(X_i)$. What can you say about$\bigcap_i X_i$? 1 Suppose, toward a contradiction, that your net$(x_i)_{i\in I}$doesn't converge to$x$. So$x$has an open neighborhood$N$such that the net doesn't ultimately get into$N$. That is, if you define$J=\{i\in I:x_i\notin N\}$, then every element of$I$is$\leq$an element of$J$. Use that to check that$(x_i)_{i\in J}\$ is a subnet of your original net ...

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