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8

You have some misconception. Open and closed are not opposites of each other. More precisely, we can have sets that are (all as subsets of $\mathbb R$ in standard topology) neiter open nor closed: $(0,1]$ both open and closed: $\mathbb R$ open and not closed: $(0,1)$ closed and not open: $[0,1]$


4

Consider the connected open set $$U=\{\,(x,y)\in\mathbb R^2\mid x^2+y^2<4, (x-1)^2+y^2>1\,\}.$$ It's closure contains $(2,0)$ and there is no polygon leading away from it: Any line segment starting there leaves the outer or enters the inner circle. (For a simply connected example, consider the upper half of $U$ only)


3

I have never seen "closed subspace" used to mean "subspace closed wrt the vector operations". This makes sense because the property of being closed wrt the vector operations is part of the meaning of the word "subspace", so it would be redundant to specify it by an adjective, while the property of being topologically closed is not redundant, and is, ...


3

Consider the application $$ f_k(M) = (\det M_{-i_1 \dots -i_k; -j_1 \dots -j_k})_{1\le i_1 <\dots < i_k\le n;1\le j_1 <\dots < j_k\le n} $$ where $ M_{-i_1 \dots -i_k; -j_1 \dots -j_k}$ is the matrix $M$ without the lines $i_a$ and without the columns $j_b$. Hence $$\text{rank }M \le r\iff M\in \bigcap_{k = r+1}^n f_k^{-1}(\{0\}) $$ Hence this ...


2

In the cofinite topology, '$A$ is closed' means '$A$ is finite or is $\mathbb{R}$' and '$A$ is open' means '$A$ has finite complement or is empty'. In particular... If $A \subseteq \mathbb{R}$ is infinite then the only closed set containing $A$ is $\mathbb{R}$, and hence $\operatorname{cl} A = \mathbb{R}$. In particular, if $A$ is infinite, every point of ...


2

It's neither open nor closed in standard topology of $\mathbb{R}$. You have argued it's not open. It's not closed since you can find a sequence in $A=(2,3]$ converging to $2$, which is not contained in $A$. "Since it is not open, it must be closed." is not correct in general. Closed is not opposite to open, "not open" is. But if you are considering ...


2

Let $S=\{2^{-n}:n\in\Bbb N\}$, and let $K=\{0\}\cup S$. Let $X=\{0,1\}\times K$. Points of $\{0,1\}\times S$ are isolated. For $i\in\{0,1\}$ let $p_i=\langle i,0\rangle$, and let $\mathscr{U}$ be a free ultrafilter on $\Bbb N$. A set $V\subseteq X$ is a nbhd of $p_i$ iff $p_i\in V$, $(\{i\}\times S)\setminus V$ is finite, and $\{n\in\Bbb N:\langle ...


2

Let $Y$ be a totally disconnected space with topology $\tau_Y$, and for each $y\in Y$ let $X_y$ be a connected space with topology $\tau_y$. Without loss of generality assume that the spaces $X_y$ are pairwise disjoint, and let $X=\bigcup_{y\in Y}X_y$. For each $y\in Y$ fix a point $p_y\in X_y$, and let $\tau_y^*=\{U\in\tau_y:p_y\notin U\}$. If $y\in ...


2

We have $\lVert f\rVert_\infty\leqslant 1$ for each $f\in D$ hence $D$ is bounded. It is equicontinuous since $$|f(x)-f(y)|\leqslant |x-y|$$ for each $x,y\in X$ and each $f\in D$.


2

Hint: $\mathbb{Q}$ is dense in $\mathbb{R}$. It suffices to show $\forall x=(x_i)\in \mathbb{R}^n$, and $x\in I=(a_1,b_1)\times (a_2,b_2)\cdots \times (a_n,b_n)\cap \mathbb{Q}^n\neq \emptyset$. That is $\exists q=(q_i)\in \mathbb{Q}^n$, s.t. $q\in I$.


1

This works fine. It's probably easier in this case to use sequential compactness. Indeed, all you have to disprove is that a single point could have an infinite preimage: but then this preimage would have an accumulation point in the projective space, which would contradict the covering property.


1

a) If $\gamma\colon S^1\rightarrow \mathbb{R}P^2$ is a generator of the fundamental group, then $S^1\times S^1\rightarrow \mathbb{R}P^2,(x,y)\mapsto \gamma(x)$ will do the job. b)/c) It holds $$[X,S^1\times S^1]=[X,S^1]\times[X,S^1]=[X,K(\mathbb{Z},1)]\times[X,K(\mathbb{Z},1)]\cong H^1(X;\mathbb{Z})\times H^1(X;\mathbb{Z})$$ for any CW-complex X. Since ...


1

Usually a continuum means either (1) connected compact Hausdorff, or (2) connected compact metric. Authors should make it clear which definition they are using. In my papers, I use the first definition, then if I use "metric continuum" to describe a space satisfying (2). The theory of non-metric continua is very interesting, so that we should not limit ...


1

$\| f\|_1 \leq C \| f \|_2$ seems obvious because a pointwise value must be less than the sup. For the other direction, my hint is that $$\|f\|_{\infty} - |f(0)| = \sup_{x \in [0,1]} |f(x)| - |f(0)| \leq \sup_{x \in [0,1]} |f(x) - f(0)|$$ Now use the fundamental theorem of calculus and some obvious inequalities.


1

Hint: $\operatorname{rank}(A)\le r$ if and only if the determinant of any $k\times k$ submatrix of $A$ vanishes for $k>r$.



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