# Tag Info

5

All of these functions are in the ball of radius $1$ centered at $0$. They don't converge uniformly to any continuous function, nor does any subsequence converge uniformly to any continuous function (why?). Hence the ball is not precompact. Conclude from there.

5

Yes, there is. Let $\langle X,\tau\rangle$ be a Hausdorff space, and let $K$ be a compact subset of $X$. Suppose that $K$ is not closed. Then we can pick $p\in(\operatorname{cl}K)\setminus K$. (Note that this does not require $\mathsf{AC}$: it’s a single choice.) Let $$\mathscr{U}=\{U\in\tau:p\notin\operatorname{cl}U\}\;.$$ ...

3

There is no such map. Other wise there'd be some $a,b\in S$ with $f(a)=\frac12$ and $f(b)=\frac32$. Then by the IVT, there exists $c$ between $a$ and $b$ (hence $c\in S$) with $f(c)=1\notin T$. The obstacle has nothing to do with compactness (and neither $S$ nor $T$ is compact), but with connectedness: $S$ is connected and $T$ is not.

3

For second question, let $m=2, n=3$. We can easily draw tree circles each pair of which intersects and all three of them have empty intersection. So the answer is negative. As for the first one, a convex set $X$ has, by definition, a property that for any two points $u, v \in X$ point $au + (1-a)v$ belongs to $X$ for any $a \in [0, 1]$. In any words, whole ...

3

In any metric space, all compact sets are closed. To see this, let $(X,d)$ be a metric space, and $Y \subset X$ a non-closed set. Since $Y$ is not closed, it does not contain all of its limit points, so there exists a point $y \not\in Y$ which is an accumulation point of $Y$. Then, the collection $$\mathscr{U} = \{U_\varepsilon\}_{\varepsilon > 0},$$ ...

3

Every open nbhd of the origin contains a set of the form $$U_\epsilon=X\cap\Big((-\epsilon,\epsilon)\times(-\epsilon,\epsilon)\Big)$$ for some $\epsilon>0$. Show that the closure of this set in $X$ is $$\operatorname{cl}_XU_\epsilon=X\cap\Big([-\epsilon,\epsilon]\times[-\epsilon,\epsilon]\Big)\;.$$ Then let $a=\dfrac{\epsilon}2$ and find a sequence ...

3

Suppose that $X$ is a metric space, and that $\langle x_n:n\in\Bbb N\rangle$ is a Cauchy sequence in $X$ that does not converge. Without loss of generality we may assume that $x_m\ne x_n$ whenever $m,n\in\Bbb N$ and $m\ne n$. Let $D=\{x_n:n\in\Bbb N\}$; then $D$ is a closed discrete set in $X$, so the map $f:D\to\Bbb R:x_n\mapsto n$ is continuous. By the ...

3

It seems the following. We shall use this question by Amathstudent and its answer by Brian M. Scott. We prove that each weak Hausdorff compactly generated $T_1$ space is $KC$. Since there is a weak Hausdorff compact $T_1$ space, which is not $KC$ (see the space $\Bbb Q^*\times\Bbb Q^*$ in the answer by Brian M. Scott), this space is not compactly ...

3

HINT: There is an $r\in\Bbb R$ such that $X\cup Y\subseteq[-r,r]$; why? What can you say about $X\times Y$ and $[-r,r]^2$? For closedness, think about the set $(\Bbb R\setminus X)\times\Bbb R$, among others. (As noted in the comments, you can use just about any characterization of compactness to prove this result without enormous difficulty; I’ve simply ...

3

The converse is not true. Consider the set of continuous functions $[0,1]\to \mathbb{R}$ in the box topology. $X$ is Hausdorff, and a sequence $x_1,x_2,\ldots$ of points of $X$ converges if and only if it is eventually constant by a diagonal argument using the fact that if two continuous functions differ at only finitely many points then they are equal. ...

3

Sure. $K = \cap_n K_n$ is non-empty and compact (closed subset of compact $K_0$, and compact sets are closed due to Hausdorffness). We know that $K \subseteq U$. If $K_m \subset U$, by decreasingness for all $k \ge m$ also $K_k \subseteq U$. So if the statement fails we know that $K_n \setminus U$ is non-empty for all $n$. The same fact that showed that ...

3

HINT: For every open set there are only finite number of $n$'s not covered by it.

2

Suppose that $$(\exists n_{0} \in \mathbb{N}) (\forall n\geq n_{0}) K_{n}\subset U$$ is not true. This means that $K_n\setminus U\ne\emptyset$ for each $n$. (Here we also use the fact that the given system is decreasing.) Then the system $(K_n\setminus U)_{n\in\mathbb N}$ is system of compact1 sets which has finite intersection property. By compactness ...

2

For each $n$ cover the space $X$ by balls of radius $1/n$. Choose finite subcovers for each $n$. Then there is at least one set in each cover with infinitely many sequence elements. Use this to create a subsequence which is Cauchy.

2

You are correct. To prove it, suppose that $K\subseteq[0,1]\times[0,1)$ is compact, and show that there is a $y\in[0,1)$ such that $K\subseteq[0,1]\times[0,y]$. This will allow you to construct a simple homeomorphism between the one-point compactification and the closed triangular region $$\{\langle x,y\rangle\in\Bbb R^2:x,y\ge 0\text{ and }x+y\le 1\}\;.$$

2

The condition says: for every $k$, the $k$th coordinate of $x\in M$ is uniformly bounded (independently of $x$) by some number $\gamma_k$. Geometrically, this means $M$ is contained in an infinite-dimensional rectangular box with sidelengths $2\gamma_1, 2\gamma_2,\dots$. The necessity of this condition follows from the fact that the projection onto $k$th ...

2

The most trivial one would be $$S_k= (0,\frac{1}{k}).$$ Note that $\bigcap_{k=1}^\infty S_k$ is empty. If only finitely many of them are not closed, then the result still hold.

2

The statement in the title is false in general: a topological space $X$ that satisfies the condition (*) Every infinite subspace $E$ of $X$ has a limit point. is called limit point compact. And a product of two spaces that are limit point compact need not be limit point compact. This shows that we cannot rely only on this property (*) to show that ...

2

I like Valeriy's answer, but let me also point out that the first question is a special case of Helly's theorem, which says that in $\mathbb R^d$, if you have at least $d+1$ sets such that any $d+1$ of them have a non-empty intersection, then the intersection of all the sets is non-empty.

2

Actually, $\{ x_A : A \in \mathcal{A} \}$ is always a zero-set in $\Psi(\mathcal{A})$: define the function $f : \Psi(\mathcal{A}) \to [0,1]$ by $$f(n) = \tfrac{1}{n+1} \\ f(x_A) = 0.$$ This is clearly a continuous function (any neighbourhood of $0$ contains all but finitely many of the $\frac{1}{n+1}$, and so it's inverse image under $f$ includes a cofinite ...

1

Hint. In a compact set in the plane every sequence has a convergent subsequence. For convenience, instead of an open ball, consider a rectangular neighborhood of $(0,0)$, $U=[-\epsilon,\epsilon]^2$ where $\epsilon>0$. Take $0<\delta<\epsilon$ and try to come up with a sequence $x_n\to 0$ such that $\delta=\sin1/{x_n}$ for all $n$. What does the ...

1

No, here is a counterexample. Define $f:[0,\infty)\to\mathbb R\times\mathbb R$ as follows: $$f(t)= \begin{cases} (t,\,-t)\text{ if }0\le t\le\frac12,\\ (t,\,t-1)\text{ if }\frac12\le t\le1,\\ (\frac1t,\,\sin^2\pi t)\text{ if }1\le t\lt\infty.\\ \end{cases}$$ The function $f$ is continuous and one-to-one. The set $S=\operatorname{range}f$ is not locally ...

1

Hint: in any metric space, any closed subset of a compact set is compact.

1

Let $\xi_n$ a sequence in $M$, we note $(\xi_n)_1$ its first coordinate, now because $|(\xi_n)_1| \leq \gamma_1$ for all $n$, by Bolzano-Weierstrass there is some subsequence $(\xi_{n_k})_1 \to \xi_1$, now take the sequence $(\xi_{n_k})$, by repeating the process there is some subsequence on the second coordinate $(\xi_{{n_k}_j})_2 \to \xi_2, we can repeat ... 1 It seems the following. The closure$\overline{Y}$of the set$Y$in the Hilbert Cube is compact, so the space$(\overline{Y},d_2|\overline{Y})$is totally bounded and its subspace$(Y,d_2|Y)$is totally bounded too. 1 Not necessarily: it could be homeomorphic to a disjoint union of any finite number of copies of$\omega+1$, for instance. In fact it can be homeomorphic to any infinite countable successor ordinal: they're all compact, zero-dimensional Hausdorff spaces. 1 It seems the following. Of course not necessarily. There is a lot even Hausodorff countable compact spaces. By Sierpinski-Mazurkiewicz Theorem, each of them is homeomorphic to$\omega^\alpha\cdot n+1$. 1 There is also another example, kind of standard, Arens space. As yet another example consider the Stone-Cech compactification of the integers$\beta \Bbb N$, and take$x$in the remainder and consider the space$X=\{x\}\cup \Bbb N$. It is known that every compact subset of$\beta \Bbb N$mist be either finite, or have cardinality$2^\frak c$, so compact ... 1 Assuming you have shown$g_{\epsilon}(x)\to g(x)$pointwise a.e.$x$, modify the estimate of$|g_{\epsilon}(x)-g(x)|$used in that proof, noting that$g$is uniformly continuous on compact sets (hint: you evoked the Lebesgue differentiation theorem (LDT) to prove a.e. convergence; with a uniformly continuous$g$, what can you say about the relevant limit in ... 1 Let$A_n = \begin{bmatrix}1&n\\0& 2 \end{bmatrix}$, we see that the eigenvalues of$A_n$are$1,2$and$A_n$is upper triangular. Since the eigenvalues are distinct,$A_n$is diagonalisable. We have$\|A_n e_2\|_2 = \sqrt{n^2+4}$, hence$\|A_n\|_2 \ge n$and so the$A_n\$ are unbounded. It follows that the sets (1), (3) are unbounded, hence are not ...

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