New answers tagged

1

Take the one-point (or any other) compactification of an uncountable discrete space. The space is open in its compactification and of course, remains discrete. Another example is $[0,\omega_1]$ with the order topology. The set of successor ordinals is uncountable and discrete.


0

Hint The eigenvalues of a matrix $A \in M_2(\Bbb R)$ are the roots of the polynomial $$\det (\lambda I_2 - A) = \lambda^2 - (\operatorname{tr} A) \lambda + \det A,$$ and these roots will be nonreal iff the discriminant $$\Delta(A) := (\operatorname{tr} A)^2 - 4 \det A$$ of $A$ is negative.


0

The usual topology is the one induced by the norm $$ \|T\|=\sup_{x\ne0}\frac{\|T(x)\|_F}{\|x\|_E} $$ for a linear map $T\in\mathcal L(E,F)$.


1

Take your favorite uncountable set $X$ and fix $x\in X$. Now consider the following topology: $$\{U\subseteq X\mid \varnothing\neq U\leftrightarrow x\in U\}$$ Namely, every non-empty open set contains $x$, and vice versa. This is not a second-countable topology, since a second-countable topology implies being Lindelöf, but $\{\{x,y\}\mid y\in X\}$ is ...


1

Let $\tau$ be the cofinite topology on $\Bbb R$. The members of $\tau$ (i.e., the open sets in this space) are $\varnothing$, the empty set, and all subsets of $\Bbb R$ of the form $\Bbb R\setminus F$, where $F$ is any finite subset of $\Bbb R$. Now let $A$ be any infinite subset of $\Bbb R$, and let $x\in\Bbb R\setminus A$. Suppose that $x\in U\in\tau$, ...


1

The interval $[0,1]$ is a compact Hausdorff space which doesn't carry the structure of a manifold without boundary. (Of course, it carries the structure of a manifold with boundary.)


1

Here is why there always exists a sequence in $K$ converging to $\sup K$, provided its existence (which is guaranteed by the fact that bounded sets always have a least upper bound in $\Bbb R$). Let $A\subseteq \Bbb R$ and let $s=\sup A<+\infty$. If there were a $n>0$ such that $\left(s-\frac1n,+\infty\right)\cap A=\emptyset$, then $s-\frac1{2n}$ ...


1

All your proofs are correct. About the fundamental group of $S^2\setminus \{(0,0,1)\}$: With the stereographic projection you can see that $S^2\setminus \{(0,0,1)\} \cong \Bbb{E}^2$ and $S^2\setminus \{(0,0,1), (0,0,-1)\} \cong \Bbb{E}^2\setminus \{0\}$. So $\pi_1(S^2\setminus \{(0,0,1)\})=\star$ and $\pi_1(S^2\setminus \{(0,0,1), (0,0,-1)\})\cong ...


0

You misquoted the book. It actually says Then $G$ is isomorphic to a group of homeomorphisms of $Y$. At least that's how it reads in my edition (copyright 1951, ninth printing, 1974).


0

Consider a connected component $C$ of $X$. (*) Given any point $x \in C$ there must exist an open set $V \subset X$ that is path-connected such that $x \in V$ We say X is locally path connected if for every point $x \in X$ and every open neighborhood $U \subset X$ of $x$, there exists a open set $V \subset X$ that is path-connected and $x \in V \subset ...


1

Following link given by Artem: $$x(t) = 0 \times 1_{t < K} + (t-K)^{3/2} \times 1_{t \ge K}$$ is a solution (check that $\lim_{t \to K} x'(t)$ exists) passes through $(t_0,0)$ for $K > t_0$


0

$\newcommand{\S}{\mathcal{S}}$ One criterion is the following: Two subbases $\S$ and $\S'$ generate the same topology iff for every $A\in\S$ and $x\in A$ there are $A_1',\dots, A_n'\in\S'$ such that $x\in A_1'\cap\dots \cap A_n'\subseteq A$, and conversely. I think you won't find something simpler.


2

Since $ [0,a) = (-2,a) \cap [0,1)$, $[0,a)$ is open in the subspace topology on $[0,1)$.


1

Yes, your proof perfectly works. Here is a related question, if you want to see. Notice that the projections are not closed in general. (For instance, the graph $G$ of $f:x\mapsto 1/x$ ($f$ being defined on $\Bbb R \setminus\{0\}$) is closed in $\Bbb R^2$ endowed with the usual topology, whereas the projection of $G$ on the $x$-axis is open, because it is ...


3

The cartesian product $[0,1]^{\Bbb R}$ is obviously compact and Hausdroff, but isn't locally homeomorphic to any $\Bbb R^n$. The really interesting (and hard) case is a topological manifold without differentiable structure.


1

Because $\langle X,\mathscr{T}\rangle$ is suborderable by $\preceq$, the topology $\mathscr{T}$ has a base $\mathscr{B}$ of $\preceq$-intervals. These intervals can be of any form: $(x,y)$, $(x,\to)$, $(\leftarrow,x)$, $(x,y]$, $[x,y)$, $[x,y]$, $[x,\to)$, or $(\leftarrow,x]$. To show that $\mathscr{T}=\mathscr{T}_{\preceq}$, it suffices to show that ...


2

Let $d$ be a metric inducing the topology of $G$ and endow $G^n$ with the metric $$d_{\infty}(x,y) = \max \,\{ d(x_k,y_k) : 1 \leqslant k \leqslant n\}.$$ Since $K^n \cap C$ is a compact subset of $B$, there is an $r > 0$ such that $B_{r}^{d_{\infty}}(K^n\cap C) \subseteq B$. The set $$C_r = C \setminus B_r^{d_{\infty}}(K^n \cap C)$$ is closed and ...


2

Any set that has an accumulation point and an isolated point


10

The union of the $x$-axis and $y$-axis in $\mathbb{R}^2$, intersected with the closed unit disk in $\mathbb{R}^2$.


1

Let $$S_1 = \biggl\{ \frac{1}{(2m+1)!} : m \in \mathbb{N}\biggr\}\quad\text{and}\quad S_2 = \{ (2n)! : n \in \mathbb{N}\}.$$ Then $S_1$ and $S_2$ are closed subsets of $\mathbb{C}^{\ast}$, $0 \in \overline{S_1}$ and $\infty \in \overline{S_2}$ (closures taken in $\mathbb{C}_{\infty}$), and yet for every compact $K \subset \mathbb{C}^{\ast}$ the set $S_1 ...


1

I'll present a fairly general way of coming up with examples of sequences which converge to more than one point. Recall that a space $X$ is called T1 (or Fréchet) if given any two distinct points $x,y \in X$ there is an open set which contains $x$ but not $y$. Suppose that $X$ is not T1. (Examples of such spaces can be found in π-Base with this search.) ...


0

This should work also: Let $X = \mathbb{R}$ with the topology $\{(-\infty, x): x \in [-\infty, \infty]\}$ $x_n \rightarrow x \:\: \Rightarrow \:\:x_n \rightarrow y \:\:\: \forall\:\: y \geq x$


3

Take $\mathbb{N}$ in the cofinite topology (the only closed sets are the finite ones (including the empty set) and $\mathbb{N}$ itself). Take $a_n$ to be any sequence where all values are different, like $a_n = n$ or $a_n = 2n$ etc.. Then $(a_n)$ converges to every point $m$ of $\mathbb{N}$, because the only open sets that contain $m$ are of the form $O = ...


4

Take the antidiscrete space with more than one point. Every sequence converges to any point in the space. Antidiscrete space $X$ has just 2 open sets: empty set and itself. Let $a_n$ be a sequence in $X$. The claim: the sequence converges to $x$ for any $x\in X$. We need to show that any open set containing $x$ contains all but finite elements of the ...


0

Stefan has given a machine for generating counterexamples in the comments. But since this question seems to have gone unanswered for so long, I figured I'd point out to provide a standard, not-so-obvious, but fairly useful instance of Stefan's general result. A somewhat stronger condition than what you've asked for is to construct an immersed submanifold ...


0

I'm going to answer my own question, and I'm madly delighted to say the answer is yes, there always is such an $f$. Note that for all $x\in X$ the set $\displaystyle N(x)=\bigcup_{O\in \mathcal{B}, x\notin \phi(O)}O$ has the form $Y\setminus\{y\}$. Indeed Suppose that $N(x)$ is all of $Y$. Then we would have $X=\phi(Y)=\phi(N(x))=\bigcup_{O\in ...


2

You can find a function $u\in C^\infty\left(\Bbb R^N\right)$ and a subsequence $u_{n_k}$ such that $u_{n_k}$ converges uniformly to $u$ on compact subsets of $\Bbb R^N$ and all the derivatives $D^\alpha u_{n_k}$ converge to $D^\alpha u$ uniformly on compact subsets. Let $E_n$ the closed ball of radius $n$. Since $\{u_k\}_{k\in\Bbb N}$ is uniformly ...


0

I have this idea that I'd love somebody with more knowledge than I will check: define a function $$\;\phi:X\to X\times X\;,\;\;\text{by}\;\;\;\phi(x):=(x,F(x))\;$$ Since each coordinate function is continous also $\;\phi\;$ is, and if $\;\Delta:=\{(x,x)\in X\times X\}\;$ is the diagonal in the cartesian product, then ...


0

$[0,1/2]$ is an infinite closed and proper subset of $X$, whereas all proper closed subsets of $Y$ are finite.


1

A space $X$ is Hausdorff iff the diagonal $\Delta$ is closed in $X \times X$. Now the set of fixed points is exactly $(F\times Id_X)^{-1}(\Delta)$ which is closed by continuity of $F$ (and thus that of $F\times Id_X$).


7

Yes, because in particular they have the same convergent sequences (a convergent sequence with its limit is a compact subset). And so the same closed sets (for metric spaces $X$, a subset $C$ is closed iff for every sequence from $C$ that converges in $X$ has its limit in $C$), and so the same open sets as well.


2

Here is an argument that is not the most general, but it works. Suppose that $f:Y\to X$ is a homeomorphism. Write $$X= \bigcup_{n≥2}[-1+1/n,1-1/n]$$ Then $$Y=f^{-1}(X) = \bigcup_{n≥2} f^{-1}([-1+1/n,1-1/n])$$ should be a countable union of finite sets, which is not possible. Notice that I only used the fact that $f$ is continuous, the fact that $f$ has a ...


3

HINT: One of the spaces is Hausdorff, and the other is not.


4

In general you can’t investigate $F(x)-x$, since the operation of subtraction may not make any sense in the space $X$. HINT: One straightforward approach is to show that the set of non-fixed points of $F$ is open. Suppose that $F(x)\ne x$. $X$ is Hausdorff, so there are open sets $U$ and $V$ such that $x\in U$, $F(x)\in V$, and $U\cap V=\varnothing$. Use ...


0

Edit: I made a mistake the first time. Here's the correct proof.


0

A very late answer: $\cal{D}(\Omega)$, the space of $C^\infty$ functions with compact support in an open $\Omega\subset \mathbb{R}^n$ with the distribution topology, seems to me (but what do I know) the most familiar space that isn't first countable. Proof that it's not first countable: Rudin Functional Analysis 6.9 tells us it's not metrizable; Rudin ...


2

So the first thing that is going on is that somewhere in your lecture notes or in your text book, they defined "map" to mean "continuous function". Now, I only know this because your quotes won't make sense if that isn't true. So I work backwards from the definitions you are providing me to work out the definitions of the terms used in your definitions. ...


1

The right space is the second! For completeness, let $\mathbb{V}$ be a complex vector space of dimension $n$; I recall that a flag of $\mathbb{V}$ is a strictly increasing sequence of vector subspaces of $\mathbb{V}$ $$ \{\underline{0}\}=\mathbb{V}_0<\mathbb{V}_1<\dots<\mathbb{V}_r\leq\mathbb{V}\,\text{where:}\,\forall ...


1

Let $A$ be a countable subset of $\Bbb R$ with at least two elements. Then $A$ is not connected. Indeed, if $a,b\in A$ with $a<b$ then $(a,b)\setminus A$ must be non-empty. If $c\in(a,b)\setminus A$ then $(-\infty,c) $ and $(c,\infty)$ are two disjoint open sets that cover $A$ and each has nontrivial intersection with $a$; hence $A$ is not connected. ...


0

There are four kinds of connected subspaces in $\Bbb R$: Empty set, if you consider it connected. Singletons, that is, sets with only one point. Intervals, that are not countable. $\Bbb R$, that is not countable.


2

In $\mathbb R$, connectedness is equivalent to path connectedness. (I was wrong about the first claim - OPEN connected $\implies$ path connected.) So if you had connected subspace with more than one point, it would contain the image of $[0,1]$, and thus be uncountable. As for your update, take something like $\mathbb Q$ and give it the trivial topology. ...


0

Taking $D = [0,1]$, I think $f(t,x) = t \sin(x/t^2)$ is a counterexample. (Here of course $f(0,x)=0$.)


0

Thanks for DanielFischer's comment. The following assumes $Y$ is Hausdorff in order to make point 5 to be true. Hints: Start with closed set $C$ in $Z$. There is an equivalent definition for continuity of $g$ with respect to closed sets. By surjective of $f$, we know $f(f^{-1}(A))=A$ for any set $A$ in $Y$. Use continuity of $g\circ f$ and $f$. Note ...


2

Suppose $T_n \rightarrow T$ and $U_n\rightarrow U$ are the convergent sequences of bounded linear maps. For any $x$, \begin{align} \|UTx - U_nT_nx\| &\leq \|(U - U_n)Tx\| + \|U_n(Tx - T_nx)\| \\ &\leq \|U-U_n\|\|Tx\| + \|U_n\|\|T-T_n\|\|x\|. \end{align} Take the limit as $n\rightarrow \infty$, the RHS tends to $0$.


0

No since $\mathbb{Q} $ is homeomorphic to $\mathbb{Q}\times \mathbb{Q} $ and $\mathbb{Q}\times \mathbb{Q} $ is not contained in any one dimensional topological manifold in $\mathbb{R}^2.$


-2

Solution in the very particular case where $X$ has a topological vector space structure: Let f-g=h which is continuous. Your set is closed because it is $h^{-1}(0) $ which the reciprocal image of closed set {0}. Edit: I have added the first sentence because of a very appropriate remark of @PtF.


3

$f-g$ is continuous again, hence $E = (f-g)^{-1}(\{0\})$ is the inverse image of a closed set with respect to a continuous map. So $E$ is closed. The Hausdorff property is used really subtle: You need it to make sure that $\{0\}$ is closed. Ok, if $X$ has no vector-space structure, we need to the following: Consider $h:Y \to X \times X, y \mapsto ...


3

$f(x+h)-f(x)-{\rm Id}\,h=0$ and so the derivative is the identity.


0

Normally the disjoint union of topological spaces $(X_\alpha)_{\alpha\in A}$ is defined as: $$X:=\bigcup_{\alpha\in A}X_{\alpha}\times\{\alpha\}$$and is accompanied by injections $\iota_{\alpha}:X_{\alpha}\to X$ prescribed by $x\mapsto\langle x,\alpha\rangle$. Preassuming that this is the case collection $\tau$ must actually be defined ...



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