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# Understand the problem

Let $X\subseteq R^2$ be the subset s.t$X=\{(x,y): x=0, |y| \leq 1\} \cup \{(x,y): 0

#### Consider the following statements:

• X is compact.
• X is connected.
• X is path-connected.

#### How many of the statements is/are true:

1. 0
2. 1
3. 2
4. 3
##### Source of the problem
TIFR 2019 GS Part A, Problem 17
Topology
Easy
##### Suggested Book
Topology, Munkres

Do you really need a hint? Try it first!

eIn general topology, a compact set K is a set for which every open cover of K contains a finite subcover of K. This is the definition of a compact set. In metric spaces, a compact space (glorified set) K is compact if it is complete and totally bounded. Completeness means that every Cauchy sequence in K converges to a point that exists in K. The rationals Q are not complete, because we can form a sequence that converges to the irrational 2–√2, which does not exist within Q. Note that complete also implies closed, because a complete space necessarily contains all of its limit points, which coincides with the properties of a closed set in metric spaces. Totally bounded means that for any fixed ε>0, K can be covered by balls of radius ε. This is different than the definition of a bounded set / space: every totally bounded set is bounded, but not every bounded set is totally bounded. Typical counter examples happen in infinite dimensions with the unit ball. We note that every metric space induces a topology because of access to the distance function, permitting a natural definition of open sets. Furthermore, every compact space still satisfies the definition of a compact set as found in general topology. In real analysis, the real numbers R form the only complete, ordered field up to isomorphism. Furthermore R induces a metric space with its own natural distance function of subtraction (accounting for signs). We gain additional properties that implies the compactness of sets with R. From Heine-Borel, a finite dimensional set $$K \subseteq \Bbb R^n$$ is compact if and only if it is closed and bounded. All previous properties of compact sets in general topology and metric spaces are still true for R. We just happen to gain convenient structure along the way. Examples: Every set F with finitely many elements is compact: leveraging the definition from general topology, for every open cover of F, it suffices to take a subcover, such that no two sets in the subcover overlap on the same points in F. This forces a finite subcover of any cover of F.
By Heine-Borel, all closed intervals of form [a,b] for some a≤b∈R are compact.
The open interval (a,b) for any a<b∈R is not compact, because it is not the case that every open cover of (a,b) has a finite subcover.

Consider the first part of the union. Is it bounded? Can you think about option 1)?

In topology and related branches of mathematics, a connected space is a topological space that cannot be represented as the union of two or more disjoint non-empty open subsets. Connectedness is one of the principal topological properties that are used to distinguish topological spaces. A subset of a topological space X is a connected set if it is a connected space when viewed as a subspace of X. Some related but stronger conditions are path connected, simply connected, and n-connected. Another related notion is locally connected, which neither implies nor follows from connectedness.

A path-connected space is a stronger notion of connectedness, requiring the structure of a path. A path from a point x to a point y in a topological space X is a continuous function ƒ from the unit interval [0,1] to X with ƒ(0) = x and ƒ(1) = y. A path-component of X is an equivalence class of X under the equivalence relation which makes x equivalent to y if there is a path from x to y. The space X is said to be path-connected (or pathwise connected or 0-connected) if there is exactly one path-component, i.e. if there is a path joining any two points in X. Again, many authors exclude empty space. Every path-connected space is connected. The converse is not always true: examples of connected spaces that are not path-connected include the extended long line L* and the topologist’s sine curve. Subsets of the real line R are connected if and only if they are path-connected; these subsets are the intervals of R. Also, open subsets of $$\Bbb R^n$$ or $$\Bbb C^n$$ are connected if and only if they are path-connected. Additionally, connectedness and path-connectedness are the same for finite topological spaces. Can you think about the second part of the union? Do you remember the example about comb space which is a famous example of connected space but not path-connected?
See this function in the second part of the union is a function of the same kind(comb space)(I will suggest you graph the function as well) but the first part of the union covers the problematic point at zero. Hence this is connected and path connected.

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