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Questions tagged [general-topology]

Everything involving general topological spaces: generation and description of topologies; open and closed sets, neighborhoods; interior, closure; connectedness; compactness; separation axioms; bases; convergence: sequences, nets and filters; continuous functions; compactifications; function spaces; etc. Please use the more specific tags, (algebraic-topology), (differential-topology), (metric-spaces), (functional-analysis) whenever appropriate.

Everything involving general topological spaces: generation and description of topologies; open and closed sets, neighborhoods; interior, closure; connectedness; compactness; precompactness; separation axioms;countability axioms; bases; convergence: sequences, nets and filters; continuous functions; compactifications; function spaces; proximity; etc. Please use the more specific tags, , , , , whenever appropriate.

57719 questions
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Are $\Bbb R^2\setminus \Bbb Q^2$ and $\Bbb R^2\setminus \Bbb Q^2\cup \{(0,0)\}$ homeomorphic?

As the title says I was wondering (being vaguely inspired by a question from Hatcher asking about the fundamental group of the first space) whether $\Bbb R^2\setminus \Bbb Q^2$ and $\Bbb R^2\setminus \Bbb Q^2\cup \{(0,0)\}$ are homeomorphic. My gut…
Alessandro Codenotti
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Continuous injective map $f:\mathbb{R}^3 \to \mathbb{R}$?

How would you show that there is no continuous injective map $f:\mathbb{R}^3 \to \mathbb{R}$? I tried the approach where if $a \in \mathbb{R}$ then $\mathbb{R} \backslash \{a\} $ is not clearly connected so there can't exist a continuous function…
user26069
25
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5 answers

Are there more general spaces than Euclidean spaces to have the Heine–Borel property?

From Wikipedia A metric space (or topological vector space) is said to have the Heine–Borel property if every closed and bounded subset is compact. Any subset of a Euclidean space, including itself, has the Heine–Borel property. I was wondering…
Tim
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Does a set $A \subseteq [0,1]$ exist such that $A$ is homeomorphic to $[0,1] \setminus A$?

Does a set $A \subseteq [0,1]$ exist such that $A$ is homeomorphic to $[0,1] \setminus A$? I have no idea how to attack this problem. Any help will be appreciated.
yes
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How is possible that those shapes are equivalent in topology?

I recently started to study topology, I have no idea about the subject so my question could be very simple but I need a clear explanation. It is about the page number 19 of Introducton to Topology by Colin Adams and Robert Franzosa; it said that…
José Marín
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24
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Closed set on Euclidean space that is not compact

I have read that a subset of Euclidean space may be called compact if it is both closed and bounded. I was wondering what a good example of a closed but unbounded set would be? Would a closed ball inside a sphere with an infinite radius do the…
ncRubert
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Possible number of open sets in a topology

Question: Is there a topological space with exactly 100 distinct open sets? My attempt : I took topological space which is countable. Maximum number of open sets in a topological space is $2^n$ which is the set of all possible subsets. Number of…
Shreya Jaganathan
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Does a topology on a countable set always have a countable base?

I'm quite new to topology, and a homework question (which I solved without knowing the answer to this question) got me thinking: If $X$ is a countable set, and $\tau$ is a topology on it, does it necessarily have a countable basis? Since $\tau…
IBS
  • 4,155
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Is whether a set is closed or not a local property?

If I want to show a topological subspace is closed in an ambient space, does it suffice to know what happens on an open cover of the ambient space? More specifically, Let $X$ is a topological space with a given open cover ${ U_i }$. Suppose that $Z…
Dalron
  • 447
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Size of the closure of a set

Why in a Hausdorff sequentially compact space the size of the closure of a countable subset is less or equal than $c$ ? I can see why this is true when the space if first countable but we are not assuming so.
student
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the equivalency of two definitions of locally closed sets

here there are 2 definitions of locally closed sets: $A$ is locally closed subset of $X$ if: a) every element in $A$ has a neighborhood $V$ in $X$ such that $A\cap V$ is closed in $V$. b) $A$ is open in its closure (in $X$) why a) and b) are…
user115608
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Open sets in product topology

For any two topological spaces $X$ and $Y$, consider $X \times Y$. Is it always true that open sets in $X \times Y$ are of the forms $U \times V$ where $U$ is open in $X$ and $V$ is open in $Y$? I think is no. Consider $\mathbb{R}^2$. Note that open…
Idonknow
  • 15,643
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Lifting local compactness in covering spaces

Since the total space of a cover is locally homeomorphic to the base space, local topological properties (like local (path) connectedness, T1 etc.) lift from the base space to the total space. The same holds for local compactness, if we assume the…
Miha Habič
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If $S \times \Bbb{R}^k$ is homeomorphic to $T \times \Bbb{R}^k$ and $S$ is compact, can we conclude that $T$ is compact?

If $S \times \Bbb{R}^k$ is homeomorphic to $T \times \Bbb{R}^k$ for some $k \geq 1$ and $S$ is compact, can we conclude that $T$ is compact? The case where $k=1$ and $S$ and hence also $T$ are locally connected is dealt with in my answer to this…
Alexander Thumm
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1 answer

If $S \times \Bbb{R}$ is homeomorphic to $T \times \Bbb{R}$ and $S$ is compact, can we conclude that $T$ is compact?

Suppose $S$ and $T$ are connected manifolds such that: 1) $S \times \mathbb{R}$ is homeomorphic to $T \times \mathbb{R}$, 2) $S$ is compact. can we conclude that $T$ is compact?
user76556
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