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So, I must prove that if I have an open cover in $M\times N$, with a finite open subcover, I must have an open cover in $M$ and $N$ with finite open subcovers. However, I don't have any idea on how to prove that. No, that is not what you must prove. What you must prove is that assuming that $M\times N$ is compact, if you have an open cover of $M$, then it ...

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No, of course not. Any finite set with more than one point is compact and not convex.

3

Take $X$ to be an uncountable set with the cocountable topology (a set is open if and only if its complement is countable). You can verify that the compact subsets of $X$ are finite sets with the discrete topology. Hence, take any $A\subseteq X$ that is not open. Then, for any compact $K\subseteq X$, $A\cap K$ is open in $K$.

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A subset of $\Bbb R^2$ is compact if and only if it is closed and bounded (Heine–Borel theorem). Here $X$ is not bounded since $(0,y)\in X$ for all $y\in\Bbb R$.

2

It’s basically correct. There’s a typo at the very beginning, where you meant to write ‘For each $n$ there exists’ (instead of $j$). And you need to pass to a convergent subsequence only once: the tail sequence $\langle x_n:n\ge k(j)\rangle$ already converges to $\hat x$, since it’s a subsequence of a sequence converging to $\hat x$. For a proof in case $X$ ...

2

I've found the answer: the two are just the same things. A comment before made me be aware that $\mathsf A$ and $\mathsf B$ can be identical, and it's not hard to prove that this is exactly the case. First, let me make some convention about terminology: A topological space $X$ is a k-space when a subspace of $X$ is closed iff the preimage along any ...

2

This is actually not about topology, but a result from general set theory. It all boils down to: $$A\setminus\bigcap B_j = \bigcup A\setminus B_j$$ Which actually is a variant of the De Morgan law.

1

A space $X$ is anticompact if and only if the only compact subsets of $X$ are the finite subsets. Let $X$ be $T_1$, anticompact, and not discrete. Since $X$ is not discrete, it has a non-empty subset $A$ that is not open. $X$ is $T_1$, so the relative topology on each finite subset of $X$ is discrete, and therefore $A\cap K$ is open in $K$ for each compact $... 1 Closedness of the graph only needs that$Y$is Hausdorff and nothing on$X$(so the assumptions are somewhat overkill for that direction). The limit$L$of$(x_n, f(x_n))$need not exist. But$(f(x_n))_n$is a sequence in a compact space$Y$, so there is a subsequence$(f(x_{n_k}))_k \rightarrow y \in Y$. Then$((x_{n_k}, f_{n_k}))_k$is a sequence in$G$, ... 1$E \ \setminus \bigcap_{k \in \mathbb{N}}\Big(\overline{\bigcup_{n \geq k}\{x_n\} }\Big)=\bigcup_{k \in \mathbb{N}}\Big(E \setminus \overline{\bigcup_{n \geq k}\{x_n\} }\Big)$is true by one of DeMorgan's Laws: https://en.wikipedia.org/wiki/De_Morgan%27s_laws#Set_theory_and_Boolean_algebra 1 Yes,$X \setminus U$must be compact. To prove this, first let's enlarge the set$K$somewhat: we may include$K$into a subset$K \subset Y$such that$Y$is a smooth, compact, surface-with-boundary. This is not too hard to see: take a collection of smooth balls whose interiors cover$K\$; by compactness finitely many of these balls suffices; we may perturb ...

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