# Tag Info

## Hot answers tagged general-topology

5

Let $\mathcal Y$ be the set of all open subsets of $X$ that are subsets of $Y$.. You have to check the three conditions: $\varnothing \in \mathcal Y$, which is clear and because $Y$ is open in $X$, and it's a subset of itself, $Y \in \mathcal Y$. The Union of arbitrarily many elements of $\mathcal Y$ is again a subset of $Y$, and each being open in $X$, ...

4

It suffices to consider the case of a product of two factors, so let $X,Y$ be nonempty topological spaces, and suppose $X\times Y$ is a $T_4$ space. Let $A,B \subset X$ be disjoint closed sets. Then $A\times Y$ and $B \times Y$ are disjoint closed sets in $X \times Y$, so there are disjoint open $U, V \subset X\times Y$ with $A\times Y \subset U$ and ...

3

Let $f:X \to Y$ be the desired open and closed continuous surjection. Since $Y$ is not $T_0$, it contains two topologically indistinguishable points. If we restrict $f$ to such two-point subset of codomain, we still get a desired continuous surjection, so we may assume that $Y$ is two-point indiscrete space. Hence, it is enough to find a $T_0$ space and its ...

2

In order to have a Suslin scheme, you must have a set $A_\sigma$ for each $\sigma\in{^{<\omega}X}$; since you have only $A_\sigma$ for $|\sigma|\le 2$, you don’t have a Suslin scheme and cannot define the result of the $\mathscr{A}$-operation. If you set $A_\sigma=\varnothing$ for all $\sigma\in{^{<\omega}X}$ of length greater than $1$, so as to have a ...

2

HINT: I would approach the union a bit differently. Let $f=f_1\cup f_2$, and suppose that $f$ is not closed; then it contains arbitrarily long arithmetic progressions. Suppose that $$\{a+kd:k=0,\ldots,\ell\}$$ is an arithmetic progession in $f$, where $\ell$ has yet to be determined (and remember that we can take $\ell$ to be as large as we want. Let ...

2

It is enough to show that when one of the norms is 1, the other is bounded between two positive constants, $m$ and $M$. It is easiest to look at the case when $||(x,y)||_\infty=1$. What does that equation tell you about $x$ and $y$? You can use this to bound $||(x,y)||_3$. In terms of the shape of $||.||_3$, you really just want to sketch the curve ...

1

Since $H_i$ is continuous we know that $H_i^{-1}(ClX_0)$ is a closed set which contains X. Since $Y_i$ is the closure of X, we know that $Y_i \subset H_i^{-1}(ClX_0)$. Therefore $H_i(Y_i) \subset ClX_0$.

1

Assume $f_n$ is a Cauchy sequence in $C([0,1])$. You said you showed that $f_n(x)$ is a Cauchy sequence of real numbers. The standard norm is complete on $\mathbb R$, so for any $x \in [0,1]$ we know $f_n(x)$ converges to some real number $\alpha_x$. Now, define a function $f$ on $[0,1]$ by $f(x)=\alpha_x$. We need to show that $f \in C([0,1])$ and that ...

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